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Neurit integrl
Robert Mak
4. bezna 2012
V tomto souboru jsou vysvtleny a na pkladech s postupnm eenmdemonstrovny zkladn integran metody. Ikonka za integrlem nateintegrl do online aplikace na adresehttp://www.mendelu.cz/user/marik/maw, kter umon vpoet integrlu,vetn automatickch nvrh, jakou metodu zvolit.
cRobert Mak, 2012
http://www.mendelu.cz/user/marik/mawhttp://www.mendelu.cz/user/marik/maw
Obsah
1 Definice neur itho integrlu 5
2 Zkladn vzorce 7
(2x + 3 4x +
6
x3 sinx + ex) dx . . . . . . . . . . . . . . . . . . . . . 8
tgx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
x + 2
x2 + 4x + 5dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
x + 5
x2 + 4dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1
(x + 6)3dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
f (ax + b) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
x + 5
x2 4x + 9dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3 Integrace per-parts 57
cRobert Mak, 2012
(x + 1) lnx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
x sinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
(x 2) sin(2x) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
(x2 + 1) sinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
(x2 + 1)ex dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
x arctgx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
lnx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
ln2 x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
x3 sinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
(x3 + 2x)ex dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
cRobert Mak, 2012
4 Integrace pomoc substituce. 125
sin(lnx)x
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
xe1x2
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
x
x4 + 16dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
ex+1
x + 1
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
tg3 x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
3x + 2 1x + 1
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167
1 +x 1x
dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
5 Dal . . . 192
arcsinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
cRobert Mak, 2012
1 Definice neur itho integrlu
Definice (neurit integrl, primitivn funkce). Bu I oteven interval, f a Ffunkce definovan na I . Jestlie plat
F (x) = f (x) pro vechna x I , (1)
nazv se funkce F primitivn funkc k funkci f , nebo t neurit integrlfunkce f na intervalu I . Zapisujeme
f (x) dx = F (x).
Existuje-li k funkci f neurit integrl na intervalu I , nazv se funkce f inte-grovateln na I .
Primitivn funkce F (x) je vdy spojit na I , plyne to z existence derivace.
Vta 1 (postaujc podmnka existence neuritho integrlu). Ke kad spojitfunkci existuje neurit integrl.
Definice neuritho integrlu cRobert Mak, 2012
Vta 2 (jednoznanost primitivn funkce). Primitivn funkce je na danm inter-valu k dan funkci urena jednoznan, a na libovolnou aditivn konstantu.Pesnji, plat nsledujc:
Je-li F primitivn funkc k funkci f na intervalu I , plat tot i pro funkci G(x) =F (x) + c, kde c R je libovoln konstanta nezvisl na x.
Jsou-li F a G primitivn funkce k te funkci f na intervalu I , li se ob funkce naintervalu I nejve o aditivn konstantu, tj. existuje c R takov, e
F (x) = G(x) + c pro vechna x I .
Bohuel, ne vdy neurit integrl dokeme efektivn najt. Zatmco problmnalezen derivace funkce sloen z funkc, kter umme derivovat, spovpouze ve sprvn aplikaci vzorc pro derivovn, problm nalzt neuritintegrl i k funkci tak jednoduch, jako je napklad ex
2
je neeiteln ve tdelementrnch funkc.
Definice neuritho integrlu cRobert Mak, 2012
2 Zkladn vzorce
Vta 3. Nech f , g jsou funkce integrovateln na I , c nech je reln slo. Pakna intervalu I plat
f (x) + g(x) dx =
f (x) dx +
g(x) dx,
cf (x) dx = c
f (x) dx.
Vta 4. Nech f je funkce integrovateln na I .
Pak
f (ax + b) dx =1aF (ax + b) , kde F je funkce primitivn k funkci f na in-
tervalu I . Plat pro ta x, pro kter je ax + b I .
Vta 5. Nech funkce f m derivaci a nem nulov bod na intervalu I . Potom
na tomto intervalu plat
f(x)
f (x)dx = ln |f (x)| .
Zkladn vzorce cRobert Mak, 2012
Najdte
(2x + 3 4x +
6
x3 sinx + ex) dx.
I =
(2x + 3 4x +
6
x3 sinx + ex) dx
= 2
x dx + 3
x14 dx + 6
x3 dx
sinx dx +
ex dx
= 2x2
2+ 3
x5/4
5/4+ 6
x2
2 ( cosx) + ex + C
= x2 +125x5/4 3 1
x2+ cosx + ex + C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?2*x+3*(x)^(1/4)+6/(x^3)-sin(x)+e^x;lang=cs
Najdte
(2x + 3 4x +
6
x3 sinx + ex) dx.
I =
(2x + 3 4x +
6
x3 sinx + ex) dx
= 2
x dx + 3
x14 dx + 6
x3 dx
sinx dx +
ex dx
= 2x2
2+ 3
x5/4
5/4+ 6
x2
2 ( cosx) + ex + C
= x2 +125x5/4 3 1
x2+ cosx + ex + C
Integrl ze soutu je souet integrl.
Integrl nsobku funkce je nsobek integrlu.
Nkter funkce je mono pepsat na mocninn funkce.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?2*x+3*(x)^(1/4)+6/(x^3)-sin(x)+e^x;lang=cs
Najdte
(2x + 3 4x +
6
x3 sinx + ex) dx.
I =
(2x + 3 4x +
6
x3 sinx + ex) dx
= 2
x dx + 3
x14 dx + 6
x3 dx
sinx dx +
ex dx
= 2x2
2+ 3
x5/4
5/4+ 6
x2
2 ( cosx) + ex + C
= x2 +125x5/4 3 1
x2+ cosx + ex + C
xn dx =xn+1
n + 1
sinx dx = cos x
ex dx = ex
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?2*x+3*(x)^(1/4)+6/(x^3)-sin(x)+e^x;lang=cs
Najdte
(2x + 3 4x +
6
x3 sinx + ex) dx.
I =
(2x + 3 4x +
6
x3 sinx + ex) dx
= 2
x dx + 3
x14 dx + 6
x3 dx
sinx dx +
ex dx
= 2x2
2+ 3
x5/4
5/4+ 6
x2
2 ( cosx) + ex + C
= x2 +125x5/4 3 1
x2+ cosx + ex + C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?2*x+3*(x)^(1/4)+6/(x^3)-sin(x)+e^x;lang=cs
Najdte
tgx dx.
I =
tgx dx
=
sinxcosx
dx
= sin x
cosxdx
=
(cosx)
cosxdx
= ln | cosx| + C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?tan(x);lang=cs
Najdte
tgx dx.
I =
tgx dx
=
sinxcosx
dx
= sin x
cosxdx
=
(cosx)
cosxdx
= ln | cosx| + C
Pouijeme definici funkce tangens. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?tan(x);lang=cs
Najdte
tgx dx.
I =
tgx dx
=
sinxcosx
dx
= sin x
cosxdx
=
(cosx)
cosxdx
= ln | cosx| + C
Plat (cosx) = sinx. itatel se tedy li od derivace jmenovatele jenomkonstantm nsobkem.
Vynsobme a vydlme integrl tmto nsobkem.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?tan(x);lang=cs
Najdte
tgx dx.
I =
tgx dx
=
sinxcosx
dx
= sin x
cosxdx
=
(cosx)
cosxdx
= ln | cosx| + C
Formln pouijeme vztah (cosx) = sinx, abychom vidli vzorec
f(x)
f (x)dx = ln |f (x)|+ C.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?tan(x);lang=cs
Najdte
tgx dx.
I =
tgx dx
=
sinxcosx
dx
= sin x
cosxdx
=
(cosx)
cosxdx
= ln | cosx| + C
f(x)
f (x)dx = ln |f (x)|+ C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?tan(x);lang=cs
Najdte
x + 2
x2 + 4x + 5dx.
I =
x + 2
x2 + 4x + 5dx
=12
2x + 4
x2 + 4x + 5dx
=12
(x2 + 4x + 5)
x2 + 4x + 5dx
=12
ln(x2 + 4x + 5) + C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+2)/(x^2+4*x+5);lang=cs
Najdte
x + 2
x2 + 4x + 5dx.
I =
x + 2
x2 + 4x + 5dx
=12
2x + 4
x2 + 4x + 5dx
=12
(x2 + 4x + 5)
x2 + 4x + 5dx
=12
ln(x2 + 4x + 5) + C
Plat (x2 + 4x + 5) = 2x + 4. itatel se tedy li od derivace jmenovatelejenom konstantm nsobkem.
Vynsobme a vydlme integrl tmto nsobkem.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+2)/(x^2+4*x+5);lang=cs
Najdte
x + 2
x2 + 4x + 5dx.
I =
x + 2
x2 + 4x + 5dx
=12
2x + 4
x2 + 4x + 5dx
=12
(x2 + 4x + 5)
x2 + 4x + 5dx
=12
ln(x2 + 4x + 5) + C
Pepeme do tvaru
f(x)
f (x)dx.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+2)/(x^2+4*x+5);lang=cs
Najdte
x + 2
x2 + 4x + 5dx.
I =
x + 2
x2 + 4x + 5dx
=12
2x + 4
x2 + 4x + 5dx
=12
(x2 + 4x + 5)
x2 + 4x + 5dx
=12
ln(x2 + 4x + 5) + C
f(x)
f (x)dx = ln |f (x)|+ C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+2)/(x^2+4*x+5);lang=cs
Najdte
x + 5
x2 + 4dx.
I =
x + 5
x2 + 4dx
=
12 2xx2 + 4
+5
x2 + 4dx
=12
ln(x2 + 4) + 512
arctgx
2+ C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2+4);lang=cs
Najdte
x + 5
x2 + 4dx.
I =
x + 5
x2 + 4dx
=
12 2xx2 + 4
+5
x2 + 4dx
=12
ln(x2 + 4) + 512
arctgx
2+ C
Derivace jmenovatele je x, v itateli vak nen nsobek tto funkce.
Vzorec
f(x)
f (x)dx nelze pmo pout.
Rozdlme zlomek na dva. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2+4);lang=cs
Najdte
x + 5
x2 + 4dx.
I =
x + 5
x2 + 4dx
=
12 2xx2 + 4
+5
x2 + 4dx
=12
ln(x2 + 4) + 512
arctgx
2+ C
V prvnm zlomku je v itateli polovina derivace jmenovatele.
Proto prvn zlomek vynsobme a vydlme dvma.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2+4);lang=cs
Najdte
x + 5
x2 + 4dx.
I =
x + 5
x2 + 4dx
=
12 2xx2 + 4
+5
x2 + 4dx
=12
ln(x2 + 4) + 512
arctgx
2+ C
f(x)
f (x)= ln |f (x)|+ C
1
A2 + x2dx =
1A
arctgx
A
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2+4);lang=cs
Najdte
1
(x + 6)3dx.
I =
1
(x + 6)3dx
=
(x + 6)3 dx
=(x + 6)2
2= 1
2(x + 6)2+ C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?1/(x+6)^3;lang=cs
Najdte
1
(x + 6)3dx.
I =
1
(x + 6)3dx
=
(x + 6)3 dx
=(x + 6)2
2= 1
2(x + 6)2+ C
Jedn se o mocninnou funkci. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?1/(x+6)^3;lang=cs
Najdte
1
(x + 6)3dx.
I =
1
(x + 6)3dx
=
(x + 6)3 dx
=(x + 6)2
2= 1
2(x + 6)2+ C
f (ax + b) dx =1aF (ax + b), kde F je integrl z f .
V naem ppad je f (x) = x3, F (x) = x2
2 a a = 1.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?1/(x+6)^3;lang=cs
Najdte
1
(x + 6)3dx.
I =
1
(x + 6)3dx
=
(x + 6)3 dx
=(x + 6)2
2= 1
2(x + 6)2+ C
Upravme. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?1/(x+6)^3;lang=cs
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
1x
dx = ln |x|
f (ax + b) dx =1aF (ax + b), v naem ppad a = 2.
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
Pepeme na mocninnou funkci. Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
xn dx =1
n + 1xn+1
f (ax + b) dx =1aF (ax + b), v naem ppad a = 1.
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
ex dx = ex
f (ax + b) dx =1aF (ax + b), v naem ppad a = 1.
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
12x + 5
dx =12
ln |2x + 5| + C
1
(2 x)5dx =
(2 1 x)5 dx
=(2 x)4
4 11
=1
4(2 x)4+ C
ex dx = ex + C
e3x dx =13e3x + C
ex dx = ex
f (ax + b) dx =1aF (ax + b), v naem ppad a = 3.
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Upravme podle vzorce (a + b)2:
(ex + ex)2 = e2x + 2exex + e2x = e2x + 2 + e2x
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Integrujeme podle vzorc
ex dx = ex ,
1 dx = x ,
f (ax + b) dx =1aF (ax + b) , kde
f (x) dx = F (x).
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Pouijeme vzorecsin(2x) = 2 sinx cosx
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Integrujeme podle vzorc
sinx dx = cosx
a
f (ax + b) dx =1aF (ax + b) , kde
f (x) dx = F (x).
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Vzorec
sin2 x =1 cos(2x)
2
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
cosx dx = sinx
f (ax + b) =1aF (ax + b)
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
Potebujeme vydlit. K tomu je mono pevst itatel na tvar, kter pozdjiumon zkrtit. Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
x2 1 + 1x + 1
=x2 1x + 1
+1
x + 1= x 1 + 1
x + 1
Zkladn vzorce cRobert Mak, 2012
Najdte nsledujc integrly.
(ex + ex)2 dx =
(e2x + 2 + e2x) dx
=12e2x + 2x1
2e2x + C
sinx cosx dx =12
sin(2x) dx =12 12 ( cos 2x) + C
sin2 x dx =12
(
1 cos(2x))
dx =12
[
x 12
sin(2x)]
+ C
x2
x + 1dx =
x2 1 + 1x + 1
dx =
x 1 + 1x + 1
dx
=x2
2 x + ln |x + 1| + C
xn dx =1
n + 1xn+1,
1x
dx = ln |x|,
f (ax + b) dx =1af (ax + b)
Zkladn vzorce cRobert Mak, 2012
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
Zaifrujeme derivaci jmenovatele, tj. vraz (2x 4), do itatele. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
Musme upravit zlomek tak, aby se zlomky v prvnm a druhm integrlurovnaly.
K tmto pravm pouijeme jenom multiplikativn a aditivn konstanty(nenadlaj moc velkou neplechu pi integraci).
Pidnm nsobku 12
mme ve druhm zlomku v itateli vraz
12
(2x 4) = x 2. Koeficient u x je v podku.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
12
(2x 4) = x 2
12
(2x 4) + 2 = x
Nyn je v itateli jenom x. Chyb slo 5.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
12
(2x 4) = x 2
12
(2x 4) + 2 = x
12
(2x 4) + 2 + 5 = x + 5
Prvn a druh zlomek jsou stejn, nedopustili jsme se dn pravy,kter by zmnila hodnotu zlomku.
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
Rozdlme zlomek na dva. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
f(x)
f (x)= ln |f (x)|+ C
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
Doplnme na tverec ve jmenovateli druhho zlomku.
x2 4x + 9 = x2 2 2 x + 22 4 + 9 = (x 2)2 + 5
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
1
A2 + x2dx =
1A
arctgx
A, kde v naem ppad A =
5
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
f (ax + b) dx =1aF (ax + b), v naem ppad a = 1
Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
Najdte
x + 5
x2 4x + 9dx.
I =
x + 5
x2 4x + 9dx
=
12 (2x 4)+2+5x2 4x + 9
dx
=
12 2x 4x2 4x + 9
+2 + 5
x2 4x + 9dx
=12
ln |x2 4x + 9| +
7
(x 2)2 + 5dx
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5 11
=12
ln |x2 4x + 9| + 7 15
arctgx 2
5+ C
Upravme. Zkladn vzorce cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs
3 Integrace per-parts
Vta 6. Nech funkce u a v maj derivace na intervalu I . Pak plat
u(x)v (x) dx = u(x)v (x)
u(x)v (x) dx, (2)
pokud integrl na prav stran existuje.
Integrace per-parts cRobert Mak, 2012
Vta 6. Nech funkce u a v maj derivace na intervalu I . Pak plat
u(x)v (x) dx = u(x)v (x)
u(x)v (x) dx, (3)
pokud integrl na prav stran existuje.Dkaz:
(uv) = uv + uv derivace souinu
(uv) dx =
uv dx +
uv dx zintegrovn a linearita integrlu
uv =
uv dx +
uv dx integrl odstran derivaci
uv
uv dx =
uv dx algebraick prava
Integrly typick pro vpo et metodou per-parts. P (x) je polynom.
P (x)ex dx,
P (x) sin(x) dx,
P (x) cos(x) dx,
P (x)arctgx dx,
P (x)lnm x dx.
Integrace per-parts cRobert Mak, 2012
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Funkce je souinem polynomu a logaritmick funkce per-parts. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Integrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = ln x a v = x + 1. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Integrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = ln x a v = x + 1. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Integrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = ln x a v = x + 1. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Roznsobme zvorku. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Dokonme integraci. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
(x + 1) lnx dx
(x + 1) lnx dx =
u = ln x u =1x
v = x + 1 v =x
2
2+ x
= ln x
(
x2
2+ x
)
1x
(
x2
2+ x
)
dx
=
(
x2
2+ x
)
ln x (
12x + 1
)
dx
=
(
x2
2+ x
)
ln x (
12 x
2
2+ x
)
=
(
x2
2+ x
)
ln x 14x2 x + C
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs
Vypotte
x sinx dx
x sinx dxu = x u = 1
v = sinx v = cosx
= x cosx
1 ( cosx) dx
= x cosx +
cosx dx
= x cosx + sinx +C
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs
Vypotte
x sinx dx
x sinx dxu = x u = 1
v = sinx v = cosx
= x cosx
1 ( cosx) dx
= x cosx +
cosx dx
= x cosx + sinx +C
Integrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = x a v = sinx. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs
Vypotte
x sinx dx
x sinx dxu = x u = 1
v = sinx v = cosx
= x cosx
1 ( cosx) dx
= x cosx +
cosx dx
= x cosx + sinx +C
Integrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = x a v = sinx. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs
Vypotte
x sinx dx
x sinx dxu = x u = 1
v = sinx v = cosx
= x cosx
1 ( cosx) dx
= x cosx +
cosx dx
= x cosx + sinx +C
Integrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = x a v = sinx. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs
Vypotte
x sinx dx
x sinx dxu = x u = 1
v = sinx v = cosx
= x cosx
1 ( cosx) dx
= x cosx +
cosx dx
= x cosx + sinx +C
Upravme. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs
Vypotte
x sinx dx
x sinx dxu = x u = 1
v = sinx v = cosx
= x cosx
1 ( cosx) dx
= x cosx +
cosx dx
= x cosx + sinx +C
Integruje druhou st:
cosx dx = sinx
Hotovo. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + C
Funkce je souinem polynomu a sinu per-parts. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + CIntegrujeme per-parts pomoc vzorce
u v dx = u v
u v dx
pi u = x 2 a v = sin(2x). Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + C
Plat
v =
v (x) dx =
sin(2x) dx = 12
cos(2x),
protoe
sinx dx = cosx a
f (ax + b) =1aF (ax + b).
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + C
u v dx = u v
u v dx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + C
Vytkneme konstantu(
12
)
z integrlu.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + CPlat
cos(2x) dx =12
sin(2x), protoe
cosx dx = sinx a
f (ax + b) =1aF (ax + b).
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x 2) sin(2x) dx
(x 2)sin(2x) dx =u = x 2 u = 1
v = sin(2x) v = 12
cos 2x
= (x 2) (
12
cos(2x))
1 (
12
cos 2x)
dx
= 12
(x 2) cos(2x) + 12
cos 2x dx
= 12
(x 2) cos(2x) + 12 12
sin(2x) + C
= 12
(x 2) cos(2x) + 14
sin(2x) + C
Upravme. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + C
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + C
Funkce je souinem polynomu a funkce sinus.
Budeme integrovat per-parts podle vzorce
u v dx = u v
u v dx
pi volb u = (x2 + 1) a v = sinx.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + C(x2 + 1) = 2x
sinx dx = cosx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + C
u v dx = u v
u v dx
Konstantn nsobek 2 a znamnko minus dme ped integrl. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + C
Jet jednou integrujeme per-parts. Nyn u = x a v = cosx. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + Cx = 1
cosx dx = sinx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + Cu v dx = u v
u v dx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + CIntegrujeme sinus:
sinx dx = cosx Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1) sinx dx
(x2 + 1) sinx dxu = x2 + 1 u = 2x
v = sinx v = cosx
= (x2 + 1) cosx + 2
x cosx dx
u = x u = 1
v = cosx v = sinx
= (x2 + 1) cosx + 2(
x sinx
sinx dx)
= (x2 + 1) cosx + 2(
x sinx ( cosx))
= (1 x2) cosx + 2x sinx + C
Upravme. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,
Integruje souin polynomu a exponenciln funkce. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,
Integrujeme per-parts.
Polynom budeme derivovat a exponencielu integrovat.
Nezapomeme, e
ex dx = ex.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,Vzorec je
u v dx = u v
u v dx. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C, Opt polynom krt exponenciln funkce.
Opt integrujeme per-parts. Opt derivujeme polynom.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,
Vzorec pro ervenou st je
uv dx = uv
uv dx, zbytek zstane.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,
ex dx = ex
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
(x2 + 1)ex dx.
(x2 + 1)ex dxu = x2 + 1 u = 2x
v = ex v = ex
= (x2 + 1)ex + 2
xex dxu = x u = 1
v = ex v = ex
= (x2 + 1)ex + 2(
xex +
ex dx)
= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,
Vytkneme (ex). Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs
Vypotte
x arctgx dx.
x arctgx dx
u = arctgx u =1
1 + x2
v = x v =x
2
2
=x2
2arctgx 1
2
x2
1 + x2dx
=x2
2arctgx 1
2
1 11 + x2
dx
=x2
2arctgx 1
2
(
x arctgx)
+ C.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs
Vypotte
x arctgx dx.
x arctgx dx
u = arctgx u =1
1 + x2
v = x v =x
2
2
=x2
2arctgx 1
2
x2
1 + x2dx
=x2
2arctgx 1
2
1 11 + x2
dx
=x2
2arctgx 1
2
(
x arctgx)
+ C.
Jedn se o souin polynomu a funkce arkustangens. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs
Vypotte
x arctgx dx.
x arctgx dx
u = arctgx u =1
1 + x2
v = x v =x
2
2
=x2
2arctgx 1
2
x2
1 + x2dx
=x2
2arctgx 1
2
1 11 + x2
dx
=x2
2arctgx 1
2
(
x arctgx)
+ C.
Budeme integrovat metodou per-parts. Budeme integrovat polynom aderivovat arkustangens. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs
Vypotte
x arctgx dx.
x arctgx dx
u = arctgx u =1
1 + x2
v = x v =x
2
2
=x2
2arctgx 1
2
x2
1 + x2dx
=x2
2arctgx 1
2
1 11 + x2
dx
=x2
2arctgx 1
2
(
x arctgx)
+ C.
uv dx = uv
uv dx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs
Vypotte
x arctgx dx.
x arctgx dx
u = arctgx u =1
1 + x2
v = x v =x
2
2
=x2
2arctgx 1
2
x2
1 + x2dx
=x2
2arctgx 1
2
1 11 + x2
dx
=x2
2arctgx 1
2
(
x arctgx)
+ C.
Musme integrovat racionln funkci. Nejprve provedeme dlen:
x2
x2 + 1=
(x2 + 1) 1x2 + 1
=x2 + 1
x2 + 1 1
x2 + 1= 1 1
x2 + 1
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs
Vypotte
x arctgx dx.
x arctgx dx
u = arctgx u =1
1 + x2
v = x v =x
2
2
=x2
2arctgx 1
2
x2
1 + x2dx
=x2
2arctgx 1
2
1 11 + x2
dx
=x2
2arctgx 1
2
(
x arctgx)
+ C.
K dokonen zbv integrovat jedniku a jeden zlomek. To provedemepomoc pslunch vzorc. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs
Vypotte
ln x dx
1 ln x dxu = lnx u =
1x
v = 1 v = x
= x lnx
1 dx
= x lnx x= x(lnx 1) + C
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs
Vypotte
ln x dx
1 ln x dxu = lnx u =
1x
v = 1 v = x
= x lnx
1 dx
= x lnx x= x(lnx 1) + C
Ve funkci je zaifrovan souin polynomu a logaritmick funkce:
1 ln x dx.
Integrujeme per-parts pi volb u = lnx a v = 1. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs
Vypotte
ln x dx
1 ln x dxu = lnx u =
1x
v = 1 v = x
= x lnx
1 dx
= x lnx x= x(lnx 1) + C
(lnx) =1x
1 dx = x
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs
Vypotte
ln x dx
1 ln x dxu = lnx u =
1x
v = 1 v = x
= x lnx
1 dx
= x lnx x= x(lnx 1) + C
u v dx = u v
u v dx
Uijeme vztah1xx = 1.
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs
Vypotte
ln x dx
1 ln x dxu = lnx u =
1x
v = 1 v = x
= x lnx
1 dx
= x lnx x= x(lnx 1) + C
1 dx = x
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs
Vypotte
ln x dx
1 ln x dxu = lnx u =
1x
v = 1 v = x
= x lnx
1 dx
= x lnx x= x(lnx 1) + C
Hotovo. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C Je zde zaifrovn souin polynomu a druh mocniny logaritmu.
Upravme funkci ln2 x na souin (1) (ln2 x) a integrujeme per-parts pivolb u = ln2 x a v = 1
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C(ln2 x) = 2 ln x(lnx) = 2 lnx
1x
1 dx = x
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C
u v dx = u v
u v dx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C
Tento trik ji znme: Napeme funkci ln x jako souin (1) lnx a integrujemeper-parts pi volb u = lnx a v = 1. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C(lnx) =
1x
1 dx = x
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C
u v dx = u v
u v dx
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C
Dopotme integrl z jedniky. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Vypotte
ln2 x dx
1 ln2 x dxu = ln2 x u =
2 ln xx
v = 1 v = x
= x ln2 x 2
ln x dx
u = ln x u =1x
v = 1 v = x
= x ln2 x 2(
x lnx
1 dx)
= x ln2 x 2(
x lnx x)
= x ln2 x 2x lnx + 2x + C
Upravme. Hotovo. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs
Najdte
x3 sinx dx.
x3 sinx dx =
u = x3 3x2 6x 6 0
v = sinx cos x sin x cosx sinx= x3 cosx (3x2 sinx) + 6x cosx 6 sinx= (x3 + 6x) cos(x) + (3x2 6) sinx + C
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x^3*sin(x);lang=cs
Najdte
x3 sinx dx.
x3 sinx dx =
u = x3 3x2 6x 6 0
v = sinx cos x sin x cosx sinx= x3 cosx (3x2 sinx) + 6x cosx 6 sinx= (x3 + 6x) cos(x) + (3x2 6) sinx + C
Tikrt integrujeme per-parts, ale vechno zapeme do jednoho sche-matu.
lut ipka reprezentuje derivovn. Derivujeme a na nulu.
erven ipka reprezentuje integrovn.
derivace derivace derivace derivace
integrace integrace integrace integrace
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x^3*sin(x);lang=cs
Najdte
x3 sinx dx.
x3 sinx dx =
u = x3 3x2 6x 6 0
v = sinx cos x sin x cosx sinx= x3 cosx (3x2 sinx) + 6x cosx 6 sinx= (x3 + 6x) cos(x) + (3x2 6) sinx + C
Nsobme ve smru ipek. Souinm ve smru lutch ipek znamnkoponechme, souinm ve smru ervench ipek znamnko zmnme avechny souiny seteme.
souin
souinsouin
souin
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x^3*sin(x);lang=cs
Najdte
x3 sinx dx.
x3 sinx dx =
u = x3 3x2 6x 6 0
v = sinx cos x sin x cosx sinx= x3 cosx (3x2 sinx) + 6x cosx 6 sinx= (x3 + 6x) cos(x) + (3x2 6) sinx + C
Upravme. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x^3*sin(x);lang=cs
Najdte
(x3 + 2x)ex dx.
(x3 + 2x)ex dx
=
u = x3+ 2x 3x2 + 2 6x 6 0
v = ex ex ex ex ex
= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex
= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8)
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^3+2x)*exp(-x);lang=cs
Najdte
(x3 + 2x)ex dx.
(x3 + 2x)ex dx
=
u = x3+ 2x 3x2 + 2 6x 6 0
v = ex ex ex ex ex
= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex
= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8) Tikrt integrujeme per-parts, ale vechno zapeme do jednoho sche-matu.
lut ipka reprezentuje derivovn. Derivujeme a na nulu.
erven ipka reprezentuje integrovn.
derivace derivace derivace derivace
integrace integrace integrace integrace
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^3+2x)*exp(-x);lang=cs
Najdte
(x3 + 2x)ex dx.
(x3 + 2x)ex dx
=
u = x3+ 2x 3x2 + 2 6x 6 0
v = ex ex ex ex ex
= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex
= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8)
Nsobme ve smru ipek. Souinm ve smru lutch ipek znamnkoponechme, souinm ve smru ervench ipek znamnko zmnme avechny souiny seteme.
souin
souinsouin
souin
Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^3+2x)*exp(-x);lang=cs
Najdte
(x3 + 2x)ex dx.
(x3 + 2x)ex dx
=
u = x3+ 2x 3x2 + 2 6x 6 0
v = ex ex ex ex ex
= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex
= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8)
Upravme. Integrace per-parts cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^3+2x)*exp(-x);lang=cs
4 Integrace pomoc substituce.
Vta 7. Nech f (t) je funkce spojit na intervalu I , nech funkce (x) m deri-vaci na intervalu J a plat (J) = I . Potom na intervalu J plat
f ((x))(x) dx =
f (t) dt, (4)
dosadme-li napravo t = (x).
Schematicky: (x) = t (x) dx = dt
Vta 8. Nech f (x) je funkce spojit na intervalu I , nech funkce (t) m nenu-lovou derivaci na intervalu J a plat (J) = I . Potom na intervalu I plat
f (x) dx =
f ((t))(t) dt, (5)
dosadme-li napravo t = 1(x), kde 1(x) je funkce inverzn k funkci (x).
Schematicky: x = (t) dx = (t) dt
Integrace pomoc substituce. cRobert Mak, 2012
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Vnitn sloka je ln x.
Derivace funkce ln x je 1x
.
Tato derivace, 1x
, je v souinu s integrovanou funkc.
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Zavedeme substituci lnx = t. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Nalezneme vztah mezi dx a dt. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Dosadme substituci. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Integrujeme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
sin(ln x)x
dx
sin(lnx)x
dx =
sin(lnx)1x
dxlnx = t
1x
dx = dt=
sin t dt
= cos t = cos(lnx) + C
Pouijeme substituci k nvratu k promnn x a pidme integran konstantu.Hotovo. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Zkusme substituovat za vnitn sloku sloen funkce e1x2
. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Hledme vztah mezi diferencily. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Derivujeme ob strany substituce. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Vyjdme odsud vraz x dx, kter figuruje uvnit integrlu. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Dosadme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Vypotte integrl pomoc vzorce. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
xe1x2
dx.
xe1x2
dx
1 x2 = t2x dx = dtx dx = 1
2dt
= 12
et dt
= 12et
= 12e1x
2
Pouijeme substituci pro nvrat k pvodn promnn. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Substituce x4+16 = t, nebo x4 = t, nejsou pln ikovn, protoe vztahmezi diferencily pi tto substituci je
4x3 dx = dt,
avak len x3 dx nikde v integrlu nen.
len x dx napovd, pout substituci x2 = t.
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Hledme vztah mezi diferencily a vyjdme z nj vraz x dx. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Substituce x2 = t vede k relaci x4 = (x2)2 = t2. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Dosadme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Uijeme vzorec
1
x2 + A2dx =
1A
arctgx
A
pi A = 4. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
x
x4 + 16dx
x
x4 + 16dx
x2 = t
2x dx = dt
x dx =12
dt
x4 = t2
=12
1
t2 + 16dt
=18
arctgt
4=
18
arctgx
2
4+ C
Uijeme zptnou substituci t = x2. Hotovo. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Vnitn sloka je
x + 1. Derivace tto vnitn sloky je
(
x + 1) =12
(x + 1)1/2 =12 1
x + 1.
Vskyt tto lene1
x + 1
uvnit integrlu (a v souinu) napovd, e provst
tuto substituci bude snadn. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Pouijeme navrenou substituci. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Najdeme vztah mezi diferencily dx a dt. Dostvme
12
1x + 1
dx = dt
a tuto relaci vynsobme slem 2. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Dosadme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Zintegrujeme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
ex+1
x + 1
dx
ex+1
x + 1
dx =
ex+1 1
x + 1dx
x + 1 = t1
2x + 1
dx = dt
1x + 1
dx = 2 dt
=
et2 dt = 2et = 2ex+1 + C
Uijeme substituci t =
x + 1 k nvratu k pvodn promnn. Hotovo. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Rozepeme funkci tgx pomoc funkc sinx a cosx.
Lich mocnina je i v itateli, i ve jmenovateli. Vybereme si tu v itateli.
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Vythneme jednu mocninu funkce sinx z itatele. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Sudou mocninu pevedeme na funkci cosx. Uijeme identitusin2 x + cos2 x = 1. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Dosadme cosx = t. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Nalezneme vztah mezi diferencily dx a dt. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Pepeme vraz sinx dx do novch promnnch. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Dosadme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Upravme Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Protoe je jmenovatel jednolenn, sta vydlit itatele vrazem t3. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Nyn integrujeme pomoc vzorc. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
tg3 x dx.
tg3 x dx =
sin3 x
cos3 xdx =
sin2 x
cos3 xsinx dx =
1 cos2xcos3x
sinx dx
cosx = t
sinx dx = dtsinx dx = dt
=
1 t2
t3dt =
t2 1t3
dt =
1t t3 dt
= ln |t| + 12t2 = ln | cosx| + 1
2 cos2 x+ C
Po integraci provedeme nvrat k pvodn promnn a pidme integrankonstantu. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C len 3x + 2 je pod odmocninou. Uijeme substituci, kter umon tuto
odmocninu odstranit.
Budeme dosazovat 3x + 2 = t2. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Nalezneme vztah mezi dx a dt. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Vyjdme dx. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Vyjdme promnnou x. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Pichystme si zptnou substituci. Vyjdme t pomoc x. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Provedeme substituci. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Upravme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Pevedeme na jeden zlomek. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + CVydlmet
2 tt2 + 1
=(t2 + 1) + (t 1)
t2 + 1= 1 +
t 1t2 + 1
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + CZskan funkce je zlomek, kter ped integrovnm rozdlme na souetzlomku, kter m v itateli derivaci jmenovatele, a zlomku, kter m v itatelijen konstantu. Oba zlomky pak snadno zintegrujeme. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + CIntegrace je ji snadn. Uijeme vztah
t
t2 + 1dt =
12
2t
t2 + 1dt =
12
ln(t2 + 1).
Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
3x + 2 1x + 1
dx.
3x + 2 1x + 1
dx
3x + 2 = t2
3 dx = 2t dt
dx =23t dt
x =13
(t2 2)
t =
3x + 2
=
t 113 (t
2 2) + 1 23t dt
= 2
t 1t2 + 1
t dt = 2
t2 tt2 + 1
dt = 2
1 +t 1t2 + 1
dt
= 2
1 tt2 + 1
1t2 + 1
dt = 2(
t 12
ln |t2 + 1| arctg t)
+ C
= 2
3x + 2 ln |3x + 3| 2 arctg
3x + 2 + C
Roznsobme zvorku a provedeme zptnou substituci pro nvrat kpromnn x. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs
Vypotte
1 +x 1x
dx.
1 +x 1x
dx =
x 1 = t2
dx = 2t dt
x = t2 + 1
x 1 = t
=
1 +t2
t2 + 1 2t dt
= 2
1 + t
t2 + 1 t dt = 2
t2 + t
t2 + 1dt = 2
1 +t 1t2 + 1
dt
= 2
1 +12 2tt2 + 1
1t2 + 1
dt
= 2[
t +12
ln |t2 + 1| arctg t]
= 2[
x 1 + 12
ln |x| arctg
x 1]
+ C
Funkce obsahuje odmocninu z linernho vrazu zavedeme substituci naodstrann odmocniny. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(1+sqrt(x-1))/x;lang=cs
Vypotte
1 +x 1x
dx.
1 +x 1x
dx =
x 1 = t2
dx = 2t dt
x = t2 + 1
x 1 = t
=
1 +t2
t2 + 1 2t dt
= 2
1 + t
t2 + 1 t dt = 2
t2 + t
t2 + 1dt = 2
1 +t 1t2 + 1
dt
= 2
1 +12 2tt2 + 1
1t2 + 1
dt
= 2[
t +12
ln |t2 + 1| arctg t]
= 2[
x 1 + 12
ln |x| arctg
x 1]
+ C
Vraz pod odmocninou je druh mocnina nov promnn. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(1+sqrt(x-1))/x;lang=cs
Vypotte
1 +x 1x
dx.
1 +x 1x
dx =
x 1 = t2
dx = 2t dt
x = t2 + 1
x 1 = t
=
1 +t2
t2 + 1 2t dt
= 2
1 + t
t2 + 1 t dt = 2
t2 + t
t2 + 1dt = 2
1 +t 1t2 + 1
dt
= 2
1 +12 2tt2 + 1
1t2 + 1
dt
= 2[
t +12
ln |t2 + 1| arctg t]
= 2[
x 1 + 12
ln |x| arctg
x 1]
+ CNalezneme vztah mezi diferencily dx a dt
(x 1) = 1 (derivace podle x)
(t2) = 2t (derivace podle t) Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(1+sqrt(x-1))/x;lang=cs
Vypotte
1 +x 1x
dx.
1 +x 1x
dx =
x 1 = t2
dx = 2t dt
x = t2 + 1
x 1 = t
=
1 +t2
t2 + 1 2t dt
= 2
1 + t
t2 + 1 t dt = 2
t2 + t
t2 + 1dt = 2
1 +t 1t2 + 1
dt
= 2
1 +12 2tt2 + 1
1t2 + 1
dt
= 2[
t +12
ln |t2 + 1| arctg t]
= 2[
x 1 + 12
ln |x| arctg
x 1]
+ C
Nalezneme x a
x 1 ze substitunho vztahu. Integrace pomoc substituce. cRobert Mak, 2012
http://wood.mendelu.cz/math/maw/integral/integralx.php?(1+sqrt(x-1))/x;lang=cs
Vypotte
1 +x 1x
dx.
1 +x 1x
dx =
x 1 = t2
dx