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8/14/2019 Chng 5a Tch phn 1
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Chapter 5: Primitive function Nguyn hm
chuong3a nick yahoo, mail: [email protected]
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33
1
mailto:[email protected]:[email protected]
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8/14/2019 Chng 5a Tch phn 1
2/35
a 1
a
xx x x ax ax
2 2
2 2
x dx1 / x dx a 1 2 / ln x
a 1 x
a
3 / a dx 4 / e dx e 5 / e .dx a.eln acosax sin ax
5 / sin ax.dx 6 / cosax.dxa a
dx tgax dx cot gax7 / 8 /
a acos ax sin ax
9 / shx.dx chx 10 / chx.dx shx
dx dx11 / coth x 12 / th x
sh x ch x
+= =
+
= = =
= =
= =
= =
= =
( )
( )( )
( ) ( )
22 2
2
2 2 2 2
2 2
dx x13 / Put x a.tan t dx a tan t 1 dt, t arctan
ax a
a tan t 1 dtdx 1 t 1 xdt .arctan C
a a a ax a a tan t 1
dx dx 1 dx dx14 / a x a x 2a a x a xa x
set a x u dx du a x v dx dv
1 dx dx 1 du
2a a x a x 2a
= = + =+
+ = = = = +
+ +
= = + + + = = + = =
+ = +
ln v ln udv 1 a x
lnu v 2a 2a a x
+ + = =
( )
( )
22 2
2
2 2 2
2 2 2 2
x x x
x x x
dx* I if x a set x a.tanh u dx a 1 tanh u .du u arg tanh x
a x
a 1 tanh u du 1 u arg tanh xI du
a a aa a .tanh udx dx arg tanh x 1 a x 1 a x
ln lna 2a a x 2a a xx a a x
e e 2ex a.tanh u a. a 1
e e e
= < = = =
= = = =
+
= = = = +
= = = + +
xa
e
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8/14/2019 Chng 5a Tch phn 1
3/35
( )
( )
22 2
2
2 2 2
2 2 2 2
x x
x x
dx* I if x a set x a.coth u dx a 1 coth u .du u arg coth x
a x
a 1 coth u du 1 u arg coth xI du
a a aa a .coth udx dx arg coth x 1 a x 1 a x 1 x a
ln ln lna 2a a x 2a a x 2a x ax a a x
e ex a.tanh u a.
e e
= > = = =
= = = =
+ = = = = = + + +
= =
x
x x2e
a 1 ae e
= + > +
2 2
2 2 2 2
2 2 2
dx x15 / I set x a.sin t with t dx a.cos t.dt, t arcsin
2 2 aa xa.cos t.dt a.cos t.dt x
I dt t arcsin Caa a .sin t a cos t
xset x a.cos t with 0 t dx a.sin t.dt, t arccos
aa.sin t.dt a.sin t.dt
Ia a .cos t a
= = = =
= = = = = +
= = =
= =
2x
dt t arccos Casin t
= = = +
( )
( ) ( ) ( )( )
2 2
2 2
2 222 2 2 2
' '2 2 2 2 2 2 2 2 2
2 2 2
2 2
2 2 22 22 2
dx16 / set x a t x
x a
t ax a t x t 2t.x x x ,
2t
t a .2t 2t t a 4t 2t 2a t adx dt dt dt
4t 2t2t
t adt
dx dt2t ln t C ln x x a Ctt ax a t
2t
+ = +
+ = = + =
+ += = =
+
= = = + = + + ++
3
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8/14/2019 Chng 5a Tch phn 1
4/35
( ) ( )
( )
( )( )
( )
2
2
2
2 2 2 2
Cch 2 second method : set x a.tan t with t x2 2
xdx a tan t 1 dt, t arctan
a
a tan t 1 dtdxI tan t 1 .dt
x a a tan t 1
dt t 1 xI ln tan C ln tg arctan C
cos t 2 4 2 a 4
= < < < < +
= + =
+ = = = ++ +
= = + + = + +
( )
( )
( )
( )
2
2 2
2
2
2
2
x.dx17 / arcsin x.dx x.arcsin x x.d arcsin x x.arcsin x
1 xu.dv u.v v.du
x.dx 1 duset x u 1 u 1 x
2 1 u1 x
arcsin x.dx x.arcsin x 1 x C
x.dx18 / arccos x.dx x.arccos x x.d arccos x x.arccos x
1 xx.dx
1 x
= =
=
= = = =
= + +
= = +
( )2 2
2
1 duset x u 1 u 1 x
2 1 u
arccos x.dx x.arccos x 1 x C
= = = =
= +
( )
( ) ( ) ( )
( )
2
2 22
2
x.dx19 / arc tgx.dx x.arc tgx x.d arc tgx x.arc tgx
1 x
x.dx 1 duset x u ln 1 u ln 1 x2 1 u1 x
arc tgx.dx x.arc tgx ln 1 x C
= = +
= = = + = +++
= + +
( )
( )( )
( )( )
2
2 2
2
2
x.dx20 / arccot x.dx x.arccot x x.d arccot x x.arctan x
1 x
x.dx 1 dudat x u ln 1 u ln 1 x
2 1 u1 x
arccot x.dx x.arccot x ln 1 x C
= = ++
= = = + = +++
= + + +
4
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8/14/2019 Chng 5a Tch phn 1
5/35
( )
( )( )
( )
( )
( ) ( )
( )( )( )
2
1
2 2 2
1 22
2 2 2
dx x 2dt* I Put t tan x 2arctan t, dx ,
sin x 2 1 t
sin x/2 2 tan x/2x x 1 2tsin x 2sin .cos 2 .
2 2 cos x/2 cos x/2 tan x/2 1 1 t
2 tan x/2 1x 1 1 1cos x 2cos 1 2 .
2 cos x/2 cos x/2 tan x/2 1
= = = =+
= = = =
+ + + = = = =
+
2
2
2 2
t
1 t
dx 2dt 2t dt xI ln t ln tan C
sin x t 21 t 1 t
+
= = = = = ++ +
2
d xdx dx dx x222 / I ln tg C
cos x 2 4sin x sin xsin x
2 22
x x xcos .d d tan
dx dx x2 2 2* I ln tan C
x x x x xsin x 22sin .cos sin .cos tan2 2 2 2 2
sin x.d* I tan x.dx
+ = = = = = + + +
= = = = = +
= =
( )
( )d cos xx
ln cos xcos x cos x
= =
( )'
2 22 2
2 2 22 22
2 2
a. sin t dtdx a a.cos t.dt a23 / I set x dx t arcsin
sin t xsin t sin tx aa.cos t.dt 1 a.cos t.dt 1 cos t.dt sin
t
I . . .costsin t sin t sin ta 1 sin t
a a
sin t sin tdt t 1 aln tg ln .arcsin
sin t 2 2 x
= = = = =
= = =
= = =
5
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8/14/2019 Chng 5a Tch phn 1
6/35
( )
( )
2 2
2 2 2 2 2
2 2
2 2 2 2 2
2
24 / Tinh calculate a x dx a x a
xDat x a sin t dx a cos tdt t arcsin
a
Vay so : a x dx a 1 sin t.a cos tdt a cos tdt1 cos 2t t
sin2t
a dt a C2 2 4
a x 2a sin t.cos t a x a sin t. 1 sin tarcsin C arcsin C
2 a 4 2 a 2
a
= = =
= =+ = = + +
= + + = + +
=
22
2 2 22x x
a . 1ax a x x a xa
arcsin C arcsin C2 a 2 2 a 2
+ + = + +
2 22 2
2 2
2 2 2 22 2 2 2
2 2 2 2
2
2 2 2 2 2 2 2 2 22 2
2 2 2 2 2 2 2 2 2 2
x.dxu x a25 / I x a dx dat du , v xdv dx x a
x .dx x a aI x. x a x. x a dx
x a x a
a .dxI x. x a x a dx x. x a I a .ln x x ax a
12I x. x a a .ln x x a I x. x a a .ln x x a
2
= += + = == +
+ = + = +
+ +
= + + + = + + + ++
= + + + + = + + + +
( )( ) n
n 1
2n n 1
n 1
n 1 n
n
n 1 n
ln a.u bd a.u bdu 1 126 / I cho give u
a.u b a a.u b a x
n.x .dx n.dxdux x
n.dxn.dx 1 axI .ln b
a aa.x b.x xbx
dx 1 b.ln a
n.ba.x b.x x
+
+
+
+
++= = = =
+ +
= =
= = = +++
= ++
6
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8/14/2019 Chng 5a Tch phn 1
7/35
2 2 2 2 2 22
n 1
n 2n n 1
n 1n 1
2 2 2n 2 n2
2n
n 1
2n 2 2
du 1 du 1 1 u 1 a.u27 / I . .arctg .arctg
b b ab ba .u b a abu a aa
1 n.x .dx n.dx
cho u dux x xn.dx
n.x .dx 1 axI .arctgaba a x .b b.x
bx
x .dx 1I .arc
nabx .a b
+
+
= = = = + +
= = =
= = =
+ +
= =
+
nb
tga.x
( ) dx28 / I ln x.dx dat set u ln x, dv dx du , v xx
dxI ln x.dx x.ln x .x x.ln x x
x
= = = = =
= = =
( )
( )
( )
( )( )
( )
'
2
2
2
u.v u.dv v.du u.dv u.v v.du
tch phan tung phan lan 2 get intergrate by parts second times
:
Choose v dy, x du v.du x.dy x.y y.dx du v d u v
u.dv u.v v.du u.v du v d u v
xVD1 Example1 : Calculate x ln x.dx ln x.d
= + =
= = = = =
= =
=
( )3 3 3
3 3 3 3
x ln x x.d ln x
3 3 3
x ln x x dx x ln x x. C3 3 x 3 9
=
= = +
1/( ) ( ) ( )
( )
( )n n 1n n 22 2 2 2 2
2 n 1 1dx xI I
2 n 1 ax a 2 n 1 a x a
= = +
+ +
Solution:
7
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8/14/2019 Chng 5a Tch phn 1
8/35
( ) ( )( )
( ) ( )
( ) ( )
'n2 2n n n2 2 2 2
n 12 2
n 12 2
2
n n n 12 2 2 2
dx 11 / Calculate I Dat u du x a .dx
x a x a
2n.x.dx2n.x x a .dx , dv dx,
x a
x x dxI 2n ,
x a x a
+
+
= = = +
+ +
= + = =+
= ++ +
( )
( )
( ) ( ) ( )
( )( )
( )( )
( ) ( )( )
( )
2 2 222 2
n n 1n 1 n 1 n n 12 2 2 2 2 2 2 2
2 2n n n 1 n 1 nn n2 2 2 2
n 1 nn 1 n 22 2 2 2 2
1 2 2
x a a dxx dx dx dx* a I a .I
x a x a x a x a
x xI 2n I a .I 2n.a .I 2n 1 .I
x a x a
dx x 2n 1I I 1
2n.ax a 2n.a x a
dx 1 xBecause I .arctga ax a
++ + +
+ +
+ +
+ = = =
+ + + +
= + = + + +
= = +
+ +
= =+
Nn (so) theo (follow) cng thc (formula) truy hi (1) c th ln lt
(in turn) tnh InSo following recurrence formula (1) we can be in
turn calculate nI
( ) ( )( )
( ) ( )( )
( )
( )( )
( ) ( )
( )
( )
( ) ( )
'1 n2 2n 1 n 1 n 1
2 2 2 2
n2 2n2 2
2 2 22 2
n 1 n 1 n n n2 2 2 2 2 2 2 2
2
n 1n 1 n2 2 2 2
dx 11 / Calculate I Dat u du x a .dx
x a x a
2x n 1du 2x 1 n x a , dv dx v x
x a
x a a dxx x dx x dxI 2 n 1 , *
x a x a x a x a
dx a .dxI a
x a x a
= = = +
+ +
= + = = =+
+ = + =+ + + +
= = + +
2 n.I
8
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9/35
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8/14/2019 Chng 5a Tch phn 1
10/35
( )
( )[ ]
( ) ( )( ) ( ) ( )( )
( ) ( )
i i in2 2
i
2n 1 2nm m
n
n nm m2 2 2 2i 1 i 1c c
m m
n nm m2 2 2i 1 i 12c c
2
m
1 c i.cPut f x , x 0,c , x , x
m mx a
c.m c.mI lim lim
i.c m.a m i.c m.a
c 1 c 1lim lim .
m mi.c m.a i.c
amm
lim f x
+ += =+ +
+ += =+ +
+
= = =+
= =
+ +
= = + +
=
( )
( )
( )
( )
m
i n 2n 12 2i 0 0
2n 3 !!dx. x
2a . 2n 2 !!x a
+
+=
= =
+
( )
( ) ( )( ) ( )( )
( )
2n 1m
n n 2n 1m 2 2 22i 1 0c , d
m . c d 2n 3 !!dx* lim 2
a . 2n 2 !!x am.d i. c d m.a
+
++ =+
= =
++ +
Solution:
( )
( )[ ]
( )
( )
( ) ( )( )
( ) ( )
i i in2 2i
2n 1m
n nm 2 2i 1c , d
m
nm 2 2i 1c , d
2
m.d i. c d1 c dPut f x , x d, c , x , x d i. x
m mx a
m . c dI lim
m.d i. c d m.a
c d 1lim
mm.d i. c d m.a
m
+ =+
+ =+
+ = = = + =
+
=
+ +
= + +
( )
( )
( ) ( )( )
( )
m
nm 2i 12
m
i n n 2n 1m 2 2 2 2i 0 0
c d 1lim
mm.d i. c d
am
2n 3 !!dx dxlim f x . x 2a . 2n 2 !!x a x a
+ =
+ +
++ =
= + +
= = = =+ +
10
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8/14/2019 Chng 5a Tch phn 1
11/35
( )( )
( ) ( )
( )
( )
2n n2
n 1n n 34 2
dx* I where 4ac b 0
ax bx c
2 n 1 12ax bI
2 n 1 .a4 n 1 .a ax bx c
+
= >+ +
+= +
+ +
Solution:
( )( )
( )
2n n
2
2 22 2
2
22 2
2
dx* I where 4ac b 0
ax bx c
bx c b 4ac bax bx c a x a x
a a 2a 4a
b 4ac bPut u x dx du, f if 4ac b 0
2a 4a
= >+ +
+ + = + + = + +
= + = = >
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
( )
n nn n n n 22 2 n 2 2 2
n n 1n n 22 2 2 2 2
1 du u 2n 3I I
a 2 n 1 .au f 2 n 1 .a u f
2 n 1 1dx xI I2 n 1 ax a 2 n 1 a x a
++
= = +
+ +
= = + + +
( ) ( ) ( ) ( )
( ) ( ) ( )
n nn n n 22 2 2
nn n 23 2
2ax bdx 2n 32aI I
2 n 1 .aax bx c 2 n 1 .a ax bx c
2ax b 2n 3
I2 n 1 .a4 n 1 .a ax bx c
+
+
+
= = ++ + + +
+ = + + +
( ) ( ) ( )
( )( )
[ ] ( )( )
[ ]
n n n2 3 2bb
2a2a
n 1 n 1b/2a b/2an 2 n 2
dx 2ax bJ
ax bx c 4 n 1 .a ax bx c
2 n 1 1 2n 3J J2 n 1 .a 2n 2 .a
++
+ + + +
+
= = + + + +
+ =
11
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8/14/2019 Chng 5a Tch phn 1
12/35
( )
( )
21 2
2 22 2
2
22 2
2
dx* I with 4ac b 0
ax bx c
bx c b 4ac bax bx c a x a x
a a 2a 4a
b 4ac bPut u x dx du, f if 4ac b 0
2a 4a
= >+ +
+ + = + + = + +
= + = = >
( )1 2 2 2 2
2 2
2 2
1 2 2 2
bxdu 1 u 1 2aI .arctan .arctan
a.f f a u f 4ac b 4ac ba
4a 4a
2ax b2 2a.arctan
4ac b 4ac b2a
dx 2 2ax bI .arctan
ax bx c4ac b 4ac b
+
= = = +
+ =
+ = = + +
( )
( )
( )[ ]
1 2 2 2 2b/2a b/2a
n n 1 b/2an n 22b/2a
2 3 44 5 4 6 5 42 2 2
dx 2 2ax b 2I . arctan .
2ax bx c 4ac b 4ac b 4ac b
2n 3dxI J
2n 2 .aax bx c
1 3 1 5 3 1I . , I . . , I . . .2.a 4a 2.a 6a 4a 2.a4ac b 4ac b
4ac b
++
+
+ +
+ = = = + +
= =
+ +
= = =
( )
( )
( )
( ) ( )2
n n 2 ! 22b/2a3!
2n 3 !!dx 2* . . with 4ac b 0
24ac bax bx c2n 2 !!.a
+
+
= >
+ +
Solution:
12
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8/14/2019 Chng 5a Tch phn 1
13/35
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8/14/2019 Chng 5a Tch phn 1
14/35
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8/14/2019 Chng 5a Tch phn 1
15/35
( )
( )( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( )
( )
'
2 2.22''
n 1 n 22 2b b2a 2aa
''2
a3 3n 2 ! n 2 !
2 22 23! 3!a
x .dx x .dxI a n 1 n n 1
ax bx c ax bx c
31 4c . 4ac b
2n 3 !! 1 4c 2n 3 !! 2. . . .2 2
4ac b 4ac b2n 2 !!.a 2n 2 !!.a
+ +
+ +
+ +
= = + + + + +
= =
=
( )
( )
( )
( )
( )
2
5n 2 !2 23!
31 4c
2n 3 !! 2. .
2 4ac b2n 2 !!.a
+
( )( ) ( ) ( )
( ) ( )
( )
( )
( )
( )( )
( ) ( )
m 2mm
n m2b2a
mm 1
mn 2 ! 2 23!
1 n m 1 ! x .dxI a
n 1 !ax bx c
2m 1 !!1 4c
2n 3 !!2. .2
4ac b . 4ac b2n 2 !!.a
+
+
+
+ =
+ +
+
=
( )
( )
( )
( )
( ) ( )
( ) ( )
( )
( )
( )
( )
( )
( ) ( )
( ) ( )
( )
( )
2m
n m2b/2a
m
mn 2 ! 2 2 m 13!
m
mn 2 ! m 2 23!
x .dx
ax bx c
2n 3 !! 4c 2m 1 !! n 1 !
. . .2 n m 1 !4ac b . 4ac b .2
2n 2 !!.a
2n 3 !! 4c . 2m 1 !! n 1 !. .
n m 1 !2 . 4ac b . 4ac b
2n 2 !!.a
+
+
+
+
+ +
+
= +
+ =
+
( )
( )
( )
( ) ( )2
n 1 3n 2 !2b/2a 2 23!
. 2n 3 !!x 2b
* .dx . with 4ac b 0ax bx c 4ac b2n!!.a
+
+ +
= >+ +
Solution:
15
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8/14/2019 Chng 5a Tch phn 1
16/35
( )
( )
( )
( )
( ) ( )
( ) ( ) ( ) ( )
2n n 2 ! 22b/2a
3!
'n ' n 1
' 2 2 2b
b/2a b/2ab
2n 3 !!dx 2I b . . with 4ac b 0
24ac bax bx c2n 2 !!.a
I b ax bx c .dx n ax bx c . ax bx c .dx
+
+
+ +
= = >
+ +
= + + = + + + +
( )
( )
( )
( )
( )
( )
( ) ( ) ( )
( )
( )
( )
( )
( )
'
n 1 n 2 ! 22b/2a b3!
1' 1
2 2 2n 2 ! b3!
3n 2 !2 23!
. 2n 3 !!x 2n .dx .
4ac bax bx c2 2n 2 !!.a
2n 3 !! 1. 2 . 4ac b . 4ac b2 2
2n 2 !!.a
12 2b
. 2n 3 !! 2.
4ac b2 2n 2 !!.a
+
+ +
+
+
= = + +
=
=
( )
( )
( )
( )
( )( )
( )
( )
n 1 3n 2 !2b/2a 2 23!
3n 2 !2 23!
. 2n 3 !!x 2b.dx .
ax bx c 4ac b2n 2n 2 !!.a
. 2n 3 !! 2b.
4ac b2n!!.a
++ +
+
=+ +
=
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
m
n m 2m 1n 2 !2b/2a 2m 23!
2
. 2n 3 !! 2m 1 !! 4a n 1 !dx* . .
n m 1 !ax bx c 4ac b2 2n 2 !!.a
with 4ac b 0
+
+ ++
+ =
+ + +
>
Solution:
16
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8/14/2019 Chng 5a Tch phn 1
17/35
( )
( )
( )
( )
( ) ( )
( ) ( ) ( ) ( )
2n n 2 ! 22b/2a
3!
'n ' n 1
' 2 2 2c
b/2a b/2ac
2n 3 !!dxI c . with 4ac b 0
4ac bax bx c2n 2 !!.a
I c ax bx c .dx n ax bx c . ax bx c .dx
+
+
+ +
= = >
+ +
= + + = + + + +
( )
( )
( )
( )
( )
( )
( ) ( ) ( )
( )
( )
( )
( )
( )
'
n 1 n 2 ! 22b/2a c3!
1' 1
2 2 2n 2 ! c3!
3n 2 !2 23!
. 2n 3 !!dx 2n .
4ac bax bx c2 2n 2 !!.a
2n 3 !! 1. 2 . 4ac b . 4ac b2 2
2n 2 !!.a
12 4a
. 2n 3 !! 2.
4ac b2 2n 2 !!.a
+
+ +
+
+
= = + +
=
=
( )
( )
( ) ( )
( )
( )
( )
( )
( )
( )
'2
''n 1 n 12 2b b
2a 2ac'
3n 2 !2 23!
c
1 n n 1 dxdxI c n
ax bx c ax bx c
12 4a
. 2n 3 !! 2.
4ac b2 2n 2 !!.a
+ +
+ +
+
+ = = + + + +
=
17
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8/14/2019 Chng 5a Tch phn 1
18/35
( )( ) ( ) ( ) ( ) ( )
( )
( ) ( )
( ) ( )
( )
( )
( )
( )( )( )
( )
mm
n m2b/2a
m
n m2b/2a
mm
2m 1n 2 !2 23!
1 n n 1 n 2 ... n m 1 dxI c
ax bx c
1 n m 1 ! dx
n 1 !ax bx c
2m 1 !!2 1 4a .
. 2n 3 !! 2.
4ac b2 2n 2 !!.a
+
+
+
+
++
+ + + =
+ +
+
= + +
+
=
( )( )
( )
( )( ) ( )
( )
( )( )
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
( )
( ) ( )
m
n m 2m 1n 2 !2b/2a 2m 23!
0
n 2.0 1n 2 !2b 20 23!2a
n 2 ! 23!
. 2n 3 !! 2m 1 !! 4a n 1 !dx . .n m 1 !
ax bx c 4ac b2 2n 2 !!.a
. 2n 3 !! 2.0 1 !! 4a n 1 !dxwhen m 0 : . .
n 0 1 !ax bx c 4ac b2 2n 2 !!.a
. 2n 3 !! 1.
4ac b2n 2 !!.a
++ ++
+
++
+
+ =+
+ +
+ = =
+ + +
=
( ) ( )( )
2m m2
n m n m2 2b/2a b/2a
x x* dx dx 0 with b 4ac 0
ax bx c ax bx c
+ +
+ +
= = >+ + + +
*
( ) ( ) ( )
( )( )
( )n n 1n n 1 22 2 2 2 2
2 n 1 1dx xI I
2 n 1 ax a 2 n 1 a x a
= = +
Solution:
( ) ( )
( )
( )
( )
( ) ( )
n n n2 2 2 2
'
n2 2 n 12 2
2n 2n n 12 2 2 2 2 2
dx 1* I Dat v , du dx
x a x a
x a dx 2nx x a dx 2nxdxdv , u x
x a x a x a
+
= = =
= = = =
18
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8/14/2019 Chng 5a Tch phn 1
19/35
( ) ( ) ( )
( )( )( )
( ) ( ) ( )
2
n n n n 12 2 2 2 2 2
2 2 22 2 2n n 1n 1 n 1 n n 12 2 2 2 2 2 2 2
dx x x dxSo : I 2n
x a x a x a
x a a dxx dx dx dxa I a .I
x a x a x a x a
+
++ + +
= = +
+= = + = +
( )( )
( )( )
( )
( )
( ) ( ) ( )
( )( )
( )
2 2n n n 1 n 1 nn n
2 2 2 2
nn 1 n 2
2 2 2
n 1n n n 1 2
2 2 2 2 2
x xI 2n I a .I 2n.a .I 2n 1 I
x a x a
2n 1 Ix
I 2n.a2n.a x a
2 n 1 1 Idx xI
2 n 1 .ax a 2 n 1 .a x a
+ +
+
= + + =
=
= =
( ) ( ) ( )
( )( )
( )
( )( )
( )[ ]
n 1n n 1 22 2 2 2 200
n 1 02
2 n 1 1 Idx x
2 n 1 .ax a 2 n 1 .a x a
2 n 1 1I
2 n 1 .a
++
+
= +
=
( )
( ) ( )( )
( )
( )
1 2 2
2 12 2 32 2 2 2 2 22 2
k
1 2 2 k 00
dx 1 a x 1 a xI ln ln
2a a x 2a a xx a
2 1dx x x 1 a xI I ln
a x2a 4a2a x a 2a x ax a
dx 1 a x ln1 ln1J lim ln 0
2a a x 2ax a
+
+
+ = = =
+
= = + = + +
= = = = +
19
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8/14/2019 Chng 5a Tch phn 1
20/35
( ) ( )( ) ( )
2n 1m
n 1 n 1m 2 2 2 2i 1 0c
c.m dx* lim 0
x ai.c m.a
++
+ ++ =+
= =
( ) ( )( )
( )
( )[ ]
n 1 nn 1 n 22 2 2 2 200
i i in 12 2i
2n 1dx xJ I 02nax a 2na x a
1 c i.cPut f x , x 0, c , x , x
m mx a
+
++ +
+
= = + =
= = =
( ) ( )( ) ( ) ( )( )
( ) ( )
( ) ( )
2n 1 2n 2m m
n 1 n 1 n 1m m2 2 2 2i 1 i 1c c
m
n 1m 2 2i 1c
2
m m
in 1 n 1m m2 2 2i 1 i 1 02c
c.m c m
I lim lim mi.c m.a i.c m.a
c 1lim
mi.c m.a
m
c 1 dx
lim lim f x . x 0m x ai.ca
m
+ +
+ + ++ += =+ +
++ =+
+
+ ++ += =+
= =
=
= = = =
( ) ( )m n2 2 2 2
0
* ln x a x a .dx 0+ =
Solution:
( )
( )( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
'
n ' '' 2 2 x xn2 20 0 n
n2 2 2 2
0
2 n2'' 2 2 2 2
0
dxI n 0 I n x a .dx a x .ln a. a
x a
ln x a x a .dx 0,
I n 1 . ln x a x a .dx 0
+ +
+
+
= = = =
= =
= =
20
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8/14/2019 Chng 5a Tch phn 1
21/35
( ) ( ) ( ) ( )
( ) ( )
( ) ( )( )
( )
m nmm 2 2 2 2
0
m n2 2 2 2
0n2 2n n n2 2
2 20 0 0 0
I 1 ln x a x a .dx 0
ln x a x a .dx 0
x a .dxI n .dn x a .dx .dn 0
ln x a
+
+
+ +
= =
=
= = =
( )
( )
( )
2n nn n n 3
2 4 2
dx 2ax b 2n 1* I with b 4ac 0 I
2n.aax bx c 4n.a ax bx c
+
+
= > = +
+ + + +
Solution:
( )( )
( )
( ) ( )
2n n2
2 22 2
2
22 22
n 1 nn 1 n 1 n n 32 2 n 3 2 2
dx* I with b 4ac 0
ax bx c
bx c b b 4acchange : ax bx c a x a x
a a 2a 4a
b b 4acPut : u x dx du, f if b 4ac 0
2a 4a
1 du u 2n 1I I
a 2n.au f 2n.a u f + + + ++
= >+ +
+ + = + + = +
= + = = >
= = +
( ) ( )
( )
n 1 nn 1 n n 32 3 2
nn n 34 2
2ax bdx 2n 12aI I
2n.aax bx c 2n.a ax bx c
2ax b 2n 1I
2n.a4n.a ax bx c
+ + +
+
+ = = + + + + +
+
= + + +
21
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8/14/2019 Chng 5a Tch phn 1
22/35
( )( )
( )( )
21 2
2 2
2 2 2
22 2
12 2 2
dx* I with b 4ac 0
ax bx c
bx c b b 4acchange : ax bx c a x a xa a 2a 4a
bPut : u x dx du,
2a
b 4ac du 1 u f f if b 4ac 0 I ln
2a.f u f 4a a u f
= >+ +
+ + = + + = +
= + =
= > = =
+
( )
22
1 2 2 2
2 2
2
2 2
2ax b b 4ac2adx 1 4aI .ln
ax bx c b 4ac 2ax b b 4ac2a
2a4a 4a
1 2ax b b 4ac.ln
b 4ac 2ax b b 4ac
+ = =
+ + + +
+ =
+ +
( )
( )
2
1 2 2 2kb/2a
2
dx 1 2a.k b b 4acJ lim ln 0
ax bx c b 4ac 2a.k b b 4ac
with b 4ac 0
+
+
+ = = =
+ + + +
>
( )( ) ( )( ) ( )
( )
n 12
m
n 1 n 1m 2 22 bi 1h
2a
2
h. 4a.m
dx* lim 0
ax bx c2a.m m.b h 2m.b m.b h 4a.m .c
with 4ac b 0
++
+ ++ =+ = =
+ + + + + +
+ +
Solution:
( )
( )( )
( ) ( ) ( ) ( ) ( )
( ) ( )( )
( )
2n2b/2a
'n ' '' 2 x x
b/2a n
2n
2 2 n2b/2a b/2a
dxI n 0 with b 4ac 0
ax bx c
I n ax bx c dx a x .ln a. a
ln ax bx c
ax bx c .ln ax bx c dx dx 0ax bx c
+
+
+ +
= = >+ +
= + + =
+ +
= + + + + = =+ +
( ) ( )( )
( )
( ) ( ) ( )( )
( )
( )
( )
2 22''
n2b/2a
m 2m
m n2b/2a
m 2
n2b/2a
ln ax bx cI n 1 dx 0,
ax bx c
ln ax bx c
I n 1 dx 0ax bx c
ln ax bx cdx 0
ax bx c
+
+
+
+ += =
+ +
+ +
= = + +
+ + =
+ +
( ) ( ) ( )
2m m2
n m n m2 2b/2a b/2a
x x
* dx dx 0 with b 4ac 0ax bx c ax bx c
+ +
+ + = = >+ + + +
Solution:
24
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8/14/2019 Chng 5a Tch phn 1
25/35
( )
( )( )
( ) ( )
( ) ( ) ( ) ( ) ( )
2n2b/2a
'n
' 2b/2a a
' n 1n n 1' ' 2 2
b/2a
dxI a 0 with b 4ac 0
ax bx c
I a ax bx c dx
u a n.u a . u a , I a n x . ax bx c dx 0,
+
+
+
= = >+ +
= + +
= = + + =
( ) ( ) ( ) ( )
( )( ) ( ) ( )
( )( )
( )
n 22'' 2.2 2
b/2a
m n mm 2m 2
b/2a
2m
n m2b/2a
I a 1 .n n 1 x . ax bx c dx 0
1 . n m 1 !I a x . ax bx c dx 0
n 1 !
xdx 0
ax bx c
+
+
+
+
= + + + =
+ = + + =
=+ +
( )( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2n2b/2a
'n' 2
b/2a b
' n 1n n 1' ' 2
b/2an 22'' 2 2
b/2a
dxI b 0 with b 4ac 0ax bx c
I b ax bx c dx
u b n.u b . u a , I a n x. ax bx c dx 0,
I a 1 .n n 1 x . ax bx c dx 0
+
+
+
+
= = >+ +
= + +
= = + + =
= + + + =
( )( ) ( ) ( )
( )( )
( )
m n mm m 2
b/2a
m
n m2b/2a
1 . n m 1 !I a x . ax bx c dx 0
n 1 !
x
dx 0ax bx c
+
+
+
+ = + + =
=+ +
25
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8/14/2019 Chng 5a Tch phn 1
26/35
( )
( )
22
2
2 2
mx n*/ I dx with 4ac b 0
ax bx c
m.ln ax bx c 2an mb 2ax b.arctg
2aa 4ac b 4ac b
+= >
+ +
+ + + = +
Solution:
( )
( )
22
2 22 2
2
22 2
2
2 2 2 2 2
mx n2 / Evaluate dx with 4ac b 0
ax bx c
bx c b 4ac bChange : ax bx c a x a x
a a 2a 4a
b 4ac bSet : u x , f if 4ac b 0
2a 4ab
m u nmx n 1 m 2udu 1 m.b2a
dx du na 2a a 2aax bx c u f u f
+ >
+ + + + = + + = + +
= + = >
+ + = = + + + + +
( ) ( )
2 2
2 2 2 2
2 2
du
u f
d u f m.ln u f m 1 m.b 1 u 1 m.b un arctg n arctg
2a a 2a f f 2a a.f 2a f u f
+
+ + = + = + +
( )
( )
2 2
2
2
2 2
2
2 2
m.ln u f mx n 1 m.b uI dx n arctg
2a a.f 2a f ax bx c
ax bx c bm.ln xa 1 2an mb 2aarctg2a 2aa 4ac b 4ac b
2a 2a
m.ln ax bx c m.ln a 2an mb 2ax b.arctg
2a a 4ac b 4ac b
++ = = + + +
+ + + = +
+ + + = +
C+
( )22 2 2
m.ln ax bx cmx n 2an mb 2ax bI dx .arctg C
2aax bx c a 4ac b 4ac b
m.lnaBecause l hang so
2a
+ + + + = = + + + +
26
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8/14/2019 Chng 5a Tch phn 1
27/35
( ) ( ) ( )
( )
( )
( )
n q q 1 q2 2 2
2
2a.n m.b .dxmx n m*/ I dx
ax bx c 2a. q 1 ax bx c 2a. ax bx c
with 4ac b 0
+
= = ++ + + + + +
>
Solution:
( )
( ) ( )
22
q 22
q2 2q2 2
2
mx n b 4ac bI dx, u x , f
2a 4aax bx c
b 4ac bax bx c a x with 4ac b 0 ,
2a 4a
+ = = + =
+ +
+ + = + + >
( ) ( ) ( )
( )
( ) ( )
( )
( )
q q q q q q2 2 2 2 2 2
2 2
q q q q2 2 2 2
q 12 2
q q 1 q2 2
bm u n
1 m 2udu 1 m.b du2aI du n
2aa 2a au f u f u f
d u fm 1 2a.n m.b du
2a2a au f u f
u fm 2a.n m.b du.
1 q2a 2a u f
+
+
+ = = + + + +
+ = + + +
+ = + +
( )
( ) ( ) ( )
q q 1 q 1 q2 2
q 1 q
q q 1 q 1 q2 2
m 1 2a.n m.b dx.
2a 2aax bx c ax bx cq 1
a am a 2a.n m.b a .dx
.2a 2aq 1 ax bx c ax bx c
+
+
= +
+ + + +
= +
+ + + +
( ) ( )
( )
( )q 1 q2 2
2a.n m.b .dxm
2a. q 1 ax bx c 2a. ax bx c
= +
+ + + +
27
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8/14/2019 Chng 5a Tch phn 1
28/35
( )
( )
22
2 2
2 2
mx n* I dx with 4ac b 0
ax bx c
m.ln ax bx c 2a.n m.b 2ax b b 4ac.ln
2a2a b 4ac 2ax b b 4ac
+=
+ +
Solution:( )
( )( )
( ) ( )
( )
( )
( )
( )( )
2
q2
q 1 q2 2
2q
2b
2a
mx n .dxI with b 4ac 0, q 1
ax bx c
2a.n m.b .dxm,
2a. q 1 ax bx c 2a. ax bx c
2a.n m.b .dx 0 with b 4ac 0
2a. ax bx c
+
+= >
+ +
= +
+ + + +
= >+ +
( )
( ) ( )
( )
( )
( ) ( )
( )
( ) ( )
q q 12 22b2a 2
b
2aq 2 q 2
q 1 q 1 q 12 2 2
mx n .dx m
ax bx c b 4ac b2a. q 1 a x2a 4a
4a. 4a .m 2. 4a .mm
4ac b 2a. q 1 4ac b q 1 4ac b2a. q 1
4a
+
+
+ =
+ + + +
= = =
29
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8/14/2019 Chng 5a Tch phn 1
30/35
Hm hu t (rational function) l t s (ratio) 2 a thc (polynomial)(
)
( )
P x
Q xnu bc (order) ca
P(x) > (greater than) order of Q(x) suy ra (derive) ( )(
)
( ) ( )( )
1P x P xH x
Q x Q x= + trong (where) bc
ca P1(x) nh hn (less than) order of Q(x)
Rational function is ratio of two polynomial( )
( )
P x
Q xif order of P(x) is greater than order of
Q(x), we can derive( )
( )( )
( )
( )1P x P x
H xQ x Q x
= + where order of P1(x) is less than degree of Q(x)
Rational fraction( )
( )
P x
Q x, where order of P(x) is less than order of Q(x), is called
standard
rational fraction.
Phn thc hu t (rational fraction)( )
( )
P x
Q xand bc ca P(x) < bc ca Q(x) gi l phn thc hu
t ng
Gi s (suppose) Q(x) khai trin c thnh (can be expanded to) tch
(product) cc tha s(factors) bc nht (first order) v bc 2 (quadratic
factors) v nghim:Suppose that Q(x) can be expanded to product of
first order factors and non-solution quadraticfactors.
( ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( )( ) ( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )
k 1 2 ko 2 k
k2 2o
1 1 2 2 k k 2 k2 2 2
2 2 31 2 4o 2 2
2o 2
P x A A AQ x a x a ...
Q x x a x a x a
Q x a x px q p 4q 0
P x M x N M x N M x N...Q x x px q x px q x px q
P x AA A AQ x a x a x b
Q x x a x bx a x b
P x A Mx NQ x a x a x px q
Q x x a x px q
= = + +
= + +
