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8/2/2019 Cac Phuong Phap Tinh Tich Phan
1/21
www.MATHVN.com CC PHNG PHP TNH TCH PHN
GV V SMinh - Email: [email protected] - www.mathvn.com 1
Chuyn 1: Cc phng php tnh tch phnCc phng php tnh tch phnCc phng php tnh tch phnCc phng php tnh tch phnThng thng ta gp cc loi tch phn sau y:
+) Loi 1: Tch phn ca hm s a thc phn thc hu t.+) Loi 2: Tch phn ca hm s cha cn thc+) Loi 3: Tch phn ca hm s lng gic
+) Loi 4: Tch phn ca hm s m v logariti vi cc tch phn c th tch theo cc phng php sau:I) Phng php bin i trc tip
Dng cc cng thc bin i v cc tch phn n gin v p dng c )a(F)b(F)x(Fdx)x(fb
a
b
a
========
+) Bin i phn thc v tng hiu cc phn thc n ginV d 1. Tnh:
1.
====
2
1
3
2
dxx
x2xI ta c 12ln)21(ln)12(ln
x
2xlndx)
x
2
x
1(I
2
1
2
1
2=++=
+==
2. ++++
====
2e
1
dx
x
4x3x2J ( ) ++=+=
+=
22e
1
2e
1
2/1 7e4e3xln4x3x4dx
x
43x2
3.
====
8
13 2
3 5
dxx3
1x3x4K =
=
=
8
1
8
1
33 423/23/1
4
207xx
4
3x
3
4dxx
3
1xx
3
4
+) Bin i nh cc cng thc lng gicV d 2. Tnh:
1.
====
2/
2/
xdx5cosx3cosI
( ) 08
x8sin
2
x2sin
2
1dxx8cosx2cos
2
12/
2/
2/
2/
=
+=+=
2.
====
2/
2/
xdx7sinx2sinJ
( )45
4
9
x9sin
5
x5sin
2
1dxx9cos)x5cos(
2
12/
2/
2/
2/
=
==
3. ====
2/
2/xdx7sinx3cosK
( ) 010x10cos
4x4cos
21dxx10sinx4sin
21xdx3cosx7sin
2/
2/
2/
2/
2/
2/
=
+=+==
4. ====
0
20 xdxcosx2sinH =
=
+=
0 0
0x4cos16
1x2cos
4
1dx
2
x2cos1x2sin hoc bin i
====
0
2 xdxcosx2sinH =
=
+=
0 0
0x4cos16
1x2cos
4
1dx
2
x2cos1x2sin
5. ++++++++++++
====
2/
6/
dxxcosxsin
x2cosx2sin1G
( ) 1xsin2xdxcos2dxxcosxsin
xsinxcos)xcosx(sin2/
6/
2/
6/
2/
6/
222
===+
++=
6.
====
2/
0
4 xdxsinE
( )16
3x2sin
4
x4sinx3
8
1dxx2cos4x4cos3
8
1dx
2
x2cos12/
0
2/
0
2/
0
2
=
+=+=
=
7. ====4/
0
2 xdxtanF
( )4
4xxtandx1
xcos
1 4/0
4/
0
2
==
= . xut: ====
2/
4/
21 xdxcotF
v
====4/
0
42 xdxtanF
+) Bin i biu thc ngoi vi phn vo trong vi phnV d 3. Tnh:
1. ++++====1
0
3 dx)1x2(I 104
)1x2(
2
1)1x2(d)1x2(
2
11
0
41
0
3=
+=++=
2. ====2
1
3 dx)1x2(1J 0
)1x2(1
41
2)1x2(
21)1x2(d)1x2(
21
1
0
2
1
0
22
1
3=
=
+==
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3. ====3/7
1
dx3x3K9
16)3x3(
9
2)3x3(d)3x3(
3
13/7
1
3
3/7
1
2/1===
4.
====
4
0 x325
dxH
3
13210)x325(
3
2)x325(d)x325(
3
11
0
2/1
4
0
2/1 =
==
5. ++++++++
====
2
1
dx1x1x
1G
3
123dx)1x1x(
2
1dx
)1x()1x(
1x1x2
1
2
1
=
+=
+
+=
(Nhn c t v mu vi bt lin hp ca mu s)
xut C)cax()bax()cb(a
1dx
caxbax
1G 331 ++++
++++++++++++
====
++++++++++++
==== vi cb;0a
6. ====1
0
dxx1xP =+=+=1
0
1
0
1
05
4dxx1)x1(dx1)1x(dxx1)11x(
7.
====
1
0
x1 xdxeQ2
= 1ee)x1(de2
1 1
0
x1
1
0
2x1 22
==
xut15
264dxx1xQ
1
0
231
====++++==== HD a x vo trong vi phn v thm bt (x2 + 1 - 1).
V d 4. Tnh:
1. 0dx)x2sin3x3cos2(I0
1 ====++++====
;41
xdxcosxsinI2/
0
32 ========
v 1exdxsineI2/
0
xcos3 ========
2. 2lnxdxtanJ4/
0
1 ========
; 2lnxdxcotJ2/
6/
2 ========
v 2ln3
2dx
xcos31
xsinJ
4/
0
3 ====++++
====
(a sinx, cosx vo trong vi phn)
3. 1cos1dxx
)xsin(lnK
e
11
========
; 2cos1dx
x
)xcos(lnK
2e
12
========
v 2dx
xln1x
1K
3e
13
====
++++
====
{a 1/x vo trong vi phn c d(lnx)}
4. ++++====3ln
1
x
x
1 dxe2
eH
e2
5lne2ln
3ln
1
x
+=+=
++++
====
2ln
0
x
x
2 dxe1
e1H =+=+
+=
2ln
0
2ln
0
x
x2ln
0
x
xx
3ln22ln3dxe1
e2dxdx
e1
e2e1
++++====2ln
0
x3 5e
dxH
7
12ln
5
15eln
5
1x
5
1
5e
dxe
5
1dx
5
1
5e
dx)e5e(
5
12ln
0
x
2ln
0
x
x2ln
0
2ln
0
x
xx
=
+=
+=
+
+=
++++
====
1
0xx
x
4 ee
dxeH
+=+=
+
=
1
0
21
0
x2
x2
x2
2
1eln
2
11eln
2
1
1e
dxe
+) Bin i nh vic xt du cc biu thc trong gi tr tuyt i tnh ====b
a
dx)m,x(fI
- Xt du hm s f(x,m) trong on [a; b] v chia [ ] ]b;c[...]c;c[]c;a[b;a n211 = trn mi on hm s f(x,m)gi mt du
- Tnh +++=b
c
c
c
c
a n
2
1
1
dx)m,x(f...dx)m,x(fdx)m,x(fI
V d 5. Tnh:
1. ++++====2
0
2 dx3x2xI Ta xt pt: 3x1x032xx 2 ===+ . Bng xt du f(x)
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Suy ra 4dx)3x2x(dx)3x2x(dx3x2xdx3x2xI2
1
2
1
0
2
2
1
2
1
0
2=+++=+++=
2.
====
1
3
3 dxxx4J tnh tng t ta c 16dxxx4dxxx4dxxx4J1
0
3
0
2
3
2
3
3=++=
3.2ln
14dx42K
3
0
x++++========
4. ====2
0
1 dxx2cos1H 22dxxsin2dxxsin2dxxsin22
0
2
0
=+==
====
0
2 dxx2sin1H
{Vit (1 sin2x) v bnh phng ca mt biu thc ri khai cn}
22dxxcosxsindxxcosxsindxxcosxsindxxcosxsin
2/3
4/3
4/3
4/
4/
00
=+++++=+=
II) Phng php i bin s
A - Phng php i bin s dng 1:
Gi s cn tnh tch phn =b
a
dx)x(fI ta thc hin cc bc sau:
- Bc 1. t x = u(t)- Bc 2. Ly vi phn dx = u(t)dt v biu th f(x)dx theo t v dt. Chng hn f(x)dx = g(t)dt- Bc 3. i cn khi x = a th u(t) = a ng vi t = ; khi x = b th u(t) = b ng vi t =
- Bc 4. Bin i =
dt)t(gI (tch phn ny d tnh hn th php i bin mi c ngha)
Cch t i bin dng 1.
Cch t 1. Nu hm s cha 2x1 th t ]2/;2/[t;tsinx ==== hoc t ];0[t;tcosx ====
V d 1. Tnh:
1.
====
1
2/2
2
2
dxx
x1A ta t ]2/;2/[t;tsinx = dx = cost.dt; i cn khi x = 2 /2 th t = 4/ ; khi x
= 1 th t = 2/ . Khi 4
4dt.
tsin
tsin1dt.
tsin
tcosdt.tcos
tsin
tsin1A
2/
4/
2
22/
4/
2
22/
4/
2
2
=
==
=
2.
====
1
02
2
dxx4
xB ta vit
=
1
02
2
dx)2/x(12
xB .
t ];0[t;tcos)2/x( = tdtsin2dxtcos2x ==
i cn suy ra ( )2
3
3
dtt2cos12tdtcos4)tdtsin2(
tcos12
)tcos2(B
2/
3/
2/
3/
2
3/
2/2
2
=+==
=
3. ====1
0
22 dxx34xC Trc ht ta vit
=
1
0
2
2 dx2
x.31x2C .
t ]2/;2/[t;tsinx2
3= a tch phn v dng:
12
1
27
32dt
2
t4cos1
33
4tdtcostsin
33
16C
3/
0
3/
0
22+=
==
Ch :
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- Nu hm s cha 0a,xa 2 >>>> th ta vit2
2
a
x1axa
= v t
=
=
];0[t;tcosa
x
]2/;2/[t;tsina
x
- Nu hm s cha 0b,a,bxa 2 >>>> th ta vit
2
2 xa
b1abxa
= v t
=
=
];0[t;tcosxa
b
]2/;2/[t;tsinxab
V d 2. Tnh:
1.
====
2
3/22
dx1xx
1E {Vit tch phn v dng 2X1 }
ta vit( )
=
2
3/2
22dx
x/11x
1E v t [ ]2/;2/t;tsin
x
1= suy ra
12dtE
3/
4/
==
2.
====
3/22
3/2
3
2
dxx
4x3G {Vit tch phn v dng 2X1 }
ta vit( )
=
3/22
3/2
3
2
dxx
x3/21.x.3G v t [ ]2/;2/t;tsin
x3
2= suy ra tch phn c dng
16
)336(3tdtcos
2
33G
3/
4/
2 +==
{Nu tch phn c dng bax 2 th vit v dng 2X1 }
Cch t 2. Nu tch phn c cha 2x1 ++++ hoc (((( ))))2x1 ++++ th ta t (((( ))))2/;2/t;ttanx ==== hoc( );0t;tcotx =
V d 3. Tnh:
1. ++++====3
3/1
2dx
x1
1M ta t ( )2/;2/t;ttanx = suy ra
6dtM
3/
6/
==
2. ++++
====
3
122
dxx1.x
1N ta t ( )2/;2/t;ttanx = suy ra
3
3218dt
.tsin
tcosN
3/
4/
2
==
3. ++++====a
0
2220a;dx
)xa(
1P
ta vit
+
=
a
0
2
24
dx
)a
x
(1a
1P v t ;ttan
a
x=
3
4/
0
2
3a4
2tdtcos
a
1P
+==
4. ++++++++====1
0
2dx
1xx
1Q
ta vit
++
=
1
0
2dx
)2
1x(
3
21
1
3
4Q v t ( )2/;2/t;ttan
2
1x
3
2=
+
9
3dt
2
3
3
4Q
1
0
==
Ch : Nu gp tch phn cha 2bxa ++++ hoc 2bxa ++++ th ta vit:
+=+
2
2x
a
b1axba hoc
2
2 x
a
b1abxa
+=+ v ta t ( )2/;2/t;ttanx
a
b=
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Cch t 3. Nu tch phn c chaxaxa
++++
hoc
xaxa
++++th ta t ta t ]2/;0[t;t2cosax ==== v lu vn dng
=+
=
tcos2t2cos1
tsin2t2cos1
2
2
V d 4. Tnh:1.
>>>>
++++====
0
a
0a;dxxaxa
I ta t ]2/;0[t;t2cosax = suy ra +
=
4/
2/
dt)t2sina2(t2cos1
t2cos1I
4
4a
=
2. ++++
====
2/2
0
dxx1
x1J ta t ]2/;0[t;t2cosx = suy ra
+=
8/
4/
dt)t2sin2(t2cos1
t2cos1J
=
4
224tdtcos4J
4/
8/
2 +==
{c th t tx1x1====
++++suy rra tch phn J v dng tch phn ca hm s hu t}
B - Phng php i bin s dng 2:
Gi s cn tnh tch phn ====b
a
dx)x(fI ta thc hin cc bc sau:
- Bc 1. t t = v(x)- Bc 2. Ly vi phn dx = u(t)dt v biu th f(x)dx theo t v dt. Chng hn f(x)dx = g(t)dt- Bc 3. i cn khi x = a th u(t) = a ng vi t = ; khi x = b th u(t) = b ng vi t =
- Bc 4. Bin i =
dt)t(gI (tch phn ny d tnh hn th php i bin mi c ngha)
Cch t i bin dng 2.Cch t 1. Nu hm s cha n mu th t t = mu s.V d 1. Tnh:
1. ====2/
0
2dx
xcos4
x2sinI
ta c th t t = 4 - cos2x suy ra3
4ln
t
dtI
4
3
==
2. ++++====4/
0
22dx
xcos2xsinx2sinJ
t xcos1xcos2xsint 222 +=+= suy ra ==2
2/34
3ln
t
dtJ
{c th h bc bin i tip mu s v cos2x sau a sin2x vo trong vi phn}
xut: ++++====2/
0
22221dx
xcosbxsina
xcosxsinJ
vi 0ba 22 >+
3. ++++====2ln
0
xdx
5e
1K ta t 5et x += 5tex = dtdxex = sau lm xut hin trong tch phn biu
thc dxex 7
12ln
5
1
t
5tln
5
1
)5t(t
dt
)5e(e
dxeK
7
6
7
6
2ln
0
xx
x
=
=
=+
=
{C th bin i trc tip7
12ln
5
1dx
5e
e
5
1dx
5e
5e
5
1dx
5e
e5e
5
1K
2ln
0
x
x2ln
0
x
x2ln
0
x
xx
=+
+
+=
+
+= }
4. ++++++++
====
2/
0
2dx
)4x2cosxsin2(
xcosx2sinH
ta t 4x2cosxsin2t += 21
2dt
t
1
2
1H
7
3
2==
{I khi khng t c MS}
5. ++++====2/
0
2
3
dxxcos1
xcosxsinG
ch rng tch m 3 = 2 +1 t
xcos1t 2+= 1txcos2 = dtxdxcosxsin2 = khi :2
2ln1)tlnt(
2
1dt
t
)1t(
2
1G
2
1
2
1
=
=
=
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6. ++=4/
0
dx2xcosxsin
x2cosM
ta t 2xcosxsint ++= dx)xsinx(cosdt = lu cos2x = (cosx+sinx)(cosx-sinx)
( ) 322
ln12tln2tt
dt)2t(
dx2xcosxsin
)xsinx)(cosxsinx(cos
M
22
3
22
3
4/
0
++==
=
++
+=
+
+
7. ++
=
4/
0
3dx
)2xcosx(sin
x2cosN
t 2xcosxsint ++= suy ra
)21(2
1
9
2
3
1
9
1
22
1
)22(
1
t
1
t
1
t
dt)2t(N
2
22
3
22
3
23+
=+
+
+
=
=
=
+ +
xut: +=4/
0
1 dx2xcosxsinx2cos
M
v +
=
4/
0
31dx
)2xcosx(sin
x2cosN
8.
Cch t 2. Nu hm s cha cn thc n )x( th t n )x(t ==== sau lu tha 2 v v ly vi phn 2 v.
V d 1. Tnh:
1. ++++++++
====
1
0
dx1x32
3x4I ta t 1x3t += ( )1t
3
1x
2= tdt
3
2dx = khi a tch phn v dng:
( )3
4ln
3
4
27
2dt
t2
6
9
2dt3t8t4
9
2dt
t2
t13t4
9
2I
2
1
2
1
2
2
1
3
=+
+=+
=
2. +
=
7
03 2
3
dxx1
xJ ta t 3 2x1t += 1tx 32 = dtt3xdx2 2=
20
141dt)tt(
2
3J
2
1
4==
3. +
=
2
1
2dx
x1x
1K ta t 2x1t += 1tx 22 = tdtxdx =
5
2
5
2
2 1t
1tln
2
1
t)1t(
tdtJ
+
=
=
4. +
=
2
13
dxx1x
1H ta t 3x1t += 1tx 23 = tdt2dxx3 2 = nhn c t v mu s vi x2 ta c:
2
12ln
3
2
1t
1tln
3
1
1t
dt
3
2
x1x
xdxH
3
2
3
2
2
2
132
+=
+
=
=
+
=
5. +
+=
3
02
35
dx1x
x2xG ta t 2x1t += 1tx 22 = tdtxdx = nhm x2.x.(x2 +2) ta c:
5
26t
5
t
t
tdt)1t)(1t(dx
1x
x.x)2x(G
2
1
52
1
223
02
22
=
=
+=
+
+=
6.
+++
=
6
13
dx1x91x9
1M ta t 6 1x9t += ( )1t9
1x
6= dtt
3
2dx
5= lu tha bc hai v bc ba
ta c:
+=
++=
+=
+= 3
2ln
6
11
3
2dt)
1t
11tt(
3
2
1t
dtt
3
2
tt
dtt
3
2M
2
1
2
2
1
32
1
23
5
V d 2. Tnh:
1. [H.2005.A] +
+=
2/
0
dxxcos31
xsinx2sinP
ta t xcos31t += )1t(3
1xcos 2 = tdt
3
2xdxsin = nhm
nhn t sinx ta c: +
+=
2/
0xcos31
xdxsin)1xcos2(P
( )27
34t
3
t2
9
2dx1t2
9
22
1
32
1
2=
+=+=
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2. dx.xsin31
x2sinx3cosQ
2
0
+
+=
ta t xsin31t += )1t(3
1xsin
2= tdt
3
2xdxcos = p dng cng thc
nhn i v nhn 3 ta vit: dx.xsin31
xcosxsin2xcos3xcos4Q
2
0
3
+
+=
xdxcos.xsin31
xsin23xsin442
0
2
+
+=
Vy +=2
1
24 dt)1t14t4(27
2Q
405
206tt
3
14t
5
4
27
22
1
35=
+=
3. [H.2006.A] +
=
2
022
dxxsin4xcos
x2sinR
ta t xsin31t 2+= )1t(3
1xsin 22 =
tdt3
2xdx2sin = . khi :
3
2t
3
2
t
tdt
3
2R
2
1
2
1
===
4.V d 3. Tnh:
1. +=
e
1
dxx
xln31xlnP
Ta t xln31t += )1t(3
1xln
2= tdt
3
2
x
dx= khi : ( )
135
116dxtt
9
2P
2
1
4==
2. +
=
e
1
dxxln21x
xln23Q
Ta t xln21t += )1t(2
1xln
2= tdt
x
dx= . Khi :
3
1139
3
tt4dt)t4(
t
tdt)1t(3Q
3
1
32
1
2
2
1
2
=
==
=
3. +
=
2ln2
2lnx 1e
dxR . Ta t 1et x += suy ra tdt2dxex =
+
+
=
=
5
3
213
13.
15
15ln
1t
dt2R
4. +
=
3
03 xe1
dxS . Ta t 3 xet = suy ra
1e
e2ln3
)1t(t
dx3S
e
1+
=+
=
5. +
=
5ln
0
x
xx
3e
dx1eeX
Cch t 3. Nu hm s cha cc i lng xsin , xcos v2
xtan th ta t
2
xtant = khi
2t1t2xsin+
= , 2
2
t1t1xcos
+
=
V d 4. Tnh:
1. dx.5xcos3xsin5
1Q
2/
0
++=
Ta t2
xtant =
2t1
dt2dx
+= v
5
8ln
3
1
4t
1tln
3
1dt
4t5t
1Q
1
0
1
0
2=
+
+=
++=
2. dx.2xcos
2
xtan
L
3/
0
+=
ta t2
xtant =
2t1
dt2dx
+= v
9
10ln3tln
3t
tdt2L
3/1
0
2
3/1
0
2=+=
+=
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3. ++=4
0
dx1x2sinx2cos
x2cosV
ta t xtant = 2t1
dtdx
+= v
++
+=
+
+=
1
0
2
1
0
2
1
0
2 )t1(2
tdt
)t1(2
dt
)t1(2
dt)t1(V
1
0
2
1 1tln4
1V ++= ta tnh
8)t1(2
dtV
ytant1
0
21
==
+= suy ra 8
2ln2dx
1x2sinx2cos
x2cosV
4
0
+=
++=
4. ++
+=
4
0
22
2
dx1xsinx2sinxcos
xtan1N
ta vit +++
=
4
0
2
dx1x2sinx2cos
xtan1N
v t
xtant = 2t1
dtdx
+= suy ra
4
2ln231tlnt
2
t
2
1dt
1t
t1
2
1N
1
0
21
0
2+
=
+++=
+
+=
5. [H.2008.B] +++
=
4
0
dx)xcosxsin1(2x2sin
4xsin
F
ta vit( )
+++
=
4
0
dx)xcosxsin1(2xcosxsin2
xcosxsin
2
1F
da
vo mi quan h gia xcosxsin + v xcosxsin ta t xcosxsint += dx)xsinx(cosdt = v
2
1txcosxsin
2
= khi
+=
+=
++=
++
=
2
1
2
1
2
2
1
222
1
22
1
1t
1
2
1
1t2t
dt
2
1
)t1(21t
dt
2
1F
Cch t 4. Da vo c im hai cn ca tch phn.
Nu tch phn c dng
=
a
a
dx)x(fI th ta c th vit +=
a
0
0
a
dx)x(fdx)x(fI t t = - x bin i
=
0
a
1 dx)x(fI
Nu tch phn c dng =
0
dx)x(fI th ta c th t t = - x
Nu tch phn c dng =2
0
dx)x(fI th ta c th t t = 2 - x
Nu tch phn c dng =
2/
0
dx)x(fI
th ta c th t t = 2
- x
Nu tch phn c dng =b
a
dx)x(fI th ta c th t t = (a + b) - x
V d 4. Tnh:
1.
=
1
1
2008 xdxsinxI ta vit +=
0
1
2008 xdxsinxI BAxdxsinx
1
0
2008+= . Ta t t = -x th A = - B. vy I = 0.
2. +=
0
2dx
xcos1
xsinxJ ta t xt = khi ++=
0
2
0
2dt
tcos1
tsintdt
tcos1
tsinJ ta i bin tip:
2dttcos1
tsin
J
2utantcos
0
21
=
====+= vJdttcos1
tsint
J
xt
0
22
=
===+=
.Vy 4JJ2J
22 ==
Cch t 4. Nu tch phn c cha 0a;cbxax 2 >++ th ta c th t cbxaxxat 2 ++= sau tnh x theo tv tnh dx theo t v dt.{Php th le}V d 5. Tnh:
1. +
=
1
02 1xx
dxI ta t 1xxxt 2 +=
1t2
t1x
2
+
= 3ln
1t2
dt2I
2
1
=
=
2. +
=
1
02 1x2x9
dxJ ta t 1x2x9x3t 2 +=
)1t3(2
1tx
2
=
2
126ln
3
1
1t3
dtJ
22
1
=
=
III)Phng php tch phn tng phn
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-Gi s cn tnh tch phn =b
a
dx)x(fI . Khi ta thc hin cc bc tnh:
Bc 1. Vit tch phn di dng: ==b
a
b
a
dx)x(h).x(gdx)x(fI
Bc 2. t
=
=
dx).x(hdv
)x(gu
=
=
dx).x(hvdx)x('gdu
Bc 3. p dng cng thc: hay =b
a
b
a
b
a
du.vv.udv.u
Cc cch t tch phn tng phn:
+Cch t 1. Nu tch phn c dng =b
a
dx.axsin).x(PI th ta s t
=
=
dx.axsindv
)x(Pu
=
=
a
axcosv
dx)x('Pdu
Nu tch phn c dng
b
a
dx.axcos).x(P th ta t
=
=
dx.axcosdv
)x(Pu
=
=
aaxsinv
dx)x('Pdu
Nu tch phn c dng b
a
ax dx.e).x(P th ta t
=
=
dx.edv
)x(Pu
ax
=
=
a
ev
dx)x('Pdu
ax
V d 5. Tnh:
1. =
0
dx.x2sin).1x3(I ta t
=
=
dx.x2sindv
1x3u
=
=
2
x2cosv
dx3du
2
3dx.x2cos
2
3
2
x2cos)1x3(I
00
=+=
2. +=2/
0
2 dx.xcos).1x(J
ta t
=
+=
dx.xcosdv
1xu 2
=
=
xsinv
xdx2du
1
2
0
2/
0
2 J24
4dx.xsin..x2xsin)1x(J
+=+=
ta tnh =2/
0
1 dx.xsin.xJ
bng cch t
=
=
dx.xsindv
xusau suy ra 1xdxcosxcosxJ
2/
0
2/
01=+=
.Vy4
42
4
4J
22
=+
=
3. +=1
0
x32 dx.e).1xx(L ta t
=
+=
dx.edv
1xxu
x3
2
1
31
0
x3
1
0
x32 L31
31edx.e).1x2(
31e)1xx(
31L
=+=
Tnh tip =1
0
x3
1 dx.e).1x2(L t
=
=
dx.edv
1x2u
x3
9
4e4L
3
1
= suy ra
27
5e5L
3
=
4. =
0
2 dx.)xsinx(M ta vit =
==
00
2
00
2 xdx2cosx2
1
4
xdx.
2
x2cos1xdx.xsinxM
xt 0dx.x2cosxMxu
xdx2cosdv0
1 ====
=
=
. vy ta c4
M2
=
5.
=
4/
0
2
dx.xsinM
ta i bin xt = a
=
2/
0
tdtsint2M
bng cch t
=
=
dt.tsindv
t2u 2M =
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+Cch t 2. Nu tch phn c dng =b
a
ax dx.bxsineI th ta t
=
=
dx.edv
bxsinu
ax
=
=
a
ev
bxdxcosbdu
ax
Nu tch phn c dng
=
b
a
ax dx.bxcoseI th ta t
=
=
dx.edv
bxcosu
ax
=
=
aev
bxdxsinbdu
ax
V d 6. Tnh:
1. =2/
0
x2dx.x3sin.eI
ta t
=
=
dxedv
x3sinu
x2
=
=
2
ev
xdx3cos3du
x2
10
x2
2/
0
x2
I2
3
2
edx.x3cose
2
3
2
ex3sinI ==
(*). Ta xt =
0
x2
1 dx.x3coseI v t
=
=
dxedv
x3cosu
x2
I2
3
2
1dx.x3sine
2
3
2
ex3cosI
0
x2
2/
0
x2
1 +=+=
thay vo (*) ta c:
+= I
2
3
2
1
2
3
2
eI
13
3e2I
+=
2. =
0
2x dx.)xsin.e(F ta vit =
=
0
x2
0
x2
0
x2 dx.x2cose2
1dx.e
2
1dx.
2
x2cos1eF
Ta xt2
1edx.e
2
1F
2
0
x2
1
==
. Sau hai ln tch phn tng phn ta tnh c4
1edx.x2cose
2
1F
2
0
x2
2
==
.
Vy ta c:8
1edx.)xsin.e(F
2
0
2x ==
+Cch t 3. Nu tch phn c dng [ ]=b
a
dx)x(Q.)x(PlnI th ta t[ ]
=
=
dx).x(Qdv
)x(Plnu
=
=
dx)x(Qv
dx)x(P
)x('Pdu
V d 7. Tnh:
1. =5
2
dx)1xln(.xI ta t[ ]
=
=
dx.xdv
1xlnu
=
=
2
xv
dx1x
1du
2 =
5
2
25
2
2
dx2x2
x)1xln(
2
xI
4
272ln48 +=
2. ++=3
0
2 dx)x1xln(J ta t
=
++=
dxdv
x1xlnu 2
=
+
=
xv
dxx1
1du
2 1)23ln(3J +=
3. =e
1
2xdxln.xK ta t
=
=
xdxdvxlnu 2 suy ra =
e
1
e
1
22xdxln.xxln
2xK . Xt =
e
1
1 xdxln.xK v t
=
=
xdxdv
xlnuth
4
1eK
4
1eK
22
1
=
+= .
4. =2
1
5dx
x
xlnH ta t
=
=
dxxdv
xlnu
5suy ra
256
2ln415dxx
4
1xln
x4
1H
e
1
5
2
14
=+=
.
5. =3/
6/
2dx
xcos
)xln(sinG
t
=
=
dxxcos
1dv
)xln(sinu
2
=
=
xtanv
xdxcotdu
=
3/
6/
3/
6/ dx)xln(sinxtanI
6
2ln343ln33 =
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6. dx)x(lnoscFe
1
=
t
=
=
dxdv
)xcos(lnu
=
=
xv
dxx
)xsin(lndu
+=
e
1
e
1dx)xsin(ln)xcos(lnxI (*). Ta xt
=
e
1
1 dx)xsin(lnF t
=
=
dxdv
)xsin(lnu
=
=
xv
dxx
)xcos(lndu
Fdx)xcos(ln)xsin(lnxFe
1
e
11==
thay
vo (*) ta c:2
1eFF1eF
+==
.
III)Phng php tm h s bt nh
A- Khi gp tch phn: ==== dx)x(Q)x(P
I vi P(x), Q(x) l cc a thc cax.
Bc 1: Nu bc ca P(x) bc ca Q(x) th ta ly P(x)chia cho Q(x) c thngA(x) v dR(x),tc l P(x) = Q(x).A(x) + R(x), vi bc R(x) < bc Q(x).
Suy ra :)x(Q)x(R
)x(A)x(Q)x(P
++++==== ++++==== dx)x(Q)x(R
dx)x(Adx)x(Q)x(P
Bc 2:
Ta i tnh : ====
dx)x(Q
)x(R
I , vi bc R(x) < bc Q(x).C th xy ra cc kh nngsau :
+Kh nng 1: Vi cbxax)x(Q 2 ++++++++==== ,( 0a ) th bc R(x) < 2 R(x) = M.x+N vcbxax
Nx.M)x(Q)x(R
2++++++++
++++====
TH1 : Q(x) c 2 nghimx1, x2, tc l: Q(x) = a(x x1)(x x2).
Chn hng sA, B sao cho:2121 xx
Bxx
A)xx)(xx(a
Nx.M)x(Q)x(R
++++
====
++++====
TH2 : Q(x) c nghim kpx0, tc l:2
0 )xx(a)x(Q ==== .
Chn hng sA, B sao cho:2
002
0 )xx(
Bxx
A
)xx(a
Nx.M)x(Q)x(R
++++
====
++++====
TH3 : Q(x) v nghim. Chn hng sA, B sao cho: B)x('Q.A)x(R ++++==== v)x(Q
B)x(Q
)x('Q.A)x(Q)x(R
++++====
+Kh nng 2: Vi dcxbxax)x(Q 23 ++++++++++++==== ,( 0a ) th bc R(x) < 3TH1: Q(x) c 3 nghim .x,x,x 321 tc l: )xx)(xx)(xx(a)x(Q 321 ====
Chn hng sA, B, Csao cho:321321 xx
Cxx
Bxx
A)xx)(xx)(xx(a
)x(R)x(Q)x(R
++++
++++
====
====
TH2: Q(x) c 1 n0 n 1x , 1 n0 kp 0x , tc l:2
01 )xx)(xx(a)x(Q ====
Chn hng sA, B, Csao cho:2
0012
01 )xx(
Cxx
Bxx
A
)xx)(xx(a
)x(R)x(Q)x(R
++++
++++
====
====
TH3: Q(x) c mt nghim0
x (bi 3), tc l: 30
)xx(a)x(Q ====
Chn hng sA, B, Csao cho:3
02
003
0 )xx(
C
)xx(
Bxx
A
)xx(a
)x(R)x(Q)x(R
++++
++++
====
====
TH4:Q(x) c ng mt nghim n 1x , tc l: )xax)(xx()x(Q2
1 ++++++++==== (trong 0a42
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1.
++++
++++++++====
0
1
2
2
dx2x3x
1xxI ta vit
+
+=
0
1
2dx
2x3x
1x41I v vit
2x
B
1x
A
2x3x
1x42
+
=
+
Sau chn c
A = -3; B = 7. Khi : ( ) 3ln72ln102xln71xln3xI0
1=+=
.
2. ++=
1
0
2 dx1xx
xJ ta vit x = A(x2 + x + 1) + B suy ra A = 1/2; B = - 1/2. Vy 21 JJJ += vi
3ln2
1
1xx
)1xx(d
2
1J
1
0
2
2
1 =++
++= v
+
+
=++
=
1
0
2
1
0
22
12
1x
3
2
dx
3
4.
2
1
1xx
dx
2
1J
Ta t utan2
1x
3
2=
+ suy ra
9
3du
3
32J
3/
6/
2
== .
3. +
=
3
1
3dx
x3x
1K ta vit
3x
cBx
x
A
x3x
123+
++=
+sau chn c A = 1/3, B = - 1/3, C = 0. V th vit c
3ln61dx
)3x(3xdx
x31K
3
1
2
3
1
=+
= {V a c x vo trong vi phn}.4.
B Khi gp tch phn ++
=
dxxcosdxsinc
xcosbxsinaI (c, d 0) th ta vit TS = A.(MS) + B.(MS) tc l chn A, B sao cho:
dcosx)'B(csinxdcosx)A(csinxbcosxasinx +++=+ hoc t2
xtant =
2t1
t2xsin
+=
2
2
t1
t1xcos
+
=
V d 1. Tnh:
1. ++
=
2/
0
dxxcosxsin
xcos5xsin3I
ta vit sinx)-B(cosxcosx)A(sinxcosx53sinx ++=+ suy ra A = 4; B = 1.
Khi : ( )
2xcosxsinlnx4xcosxsin
)xcosx(sinddx4I
2/
0
2/
0
2/
0
=++=+
++=
2. ++
=
2/
0
3dx
)xcosx(sin
xcosxsin3J
ta vit sinx)-B(cosxcosx)A(sinxcosx3sinx ++=+ suy ra A = 2; B = -1.
Khi : 2)xcosx(sin2
1)
4xcot(
)xcosx(sin
)xcosx(sinddx
)xcosx(sin
2I
2/
0
2
2/
0
3
2/
0
2=
+++=
+
+
+=
C Khi gp tch phn ++++
=
dxnxcosdxsinc
mxcosbxsinaI (c, d 0) th ta vit TS = A.(MS) + B.(MS) + C. Chn A, B,C sao cho:
Cn)'dcosxB(csinxn)dcosxA(csinxmbcosxasinx ++++++=++ hoc c th t
2
xtant =
2t1
t2xsin
+=
2
2
t1
t1xcos
+
=
V d 1. Tnh:
1. +++
=
2/
0
dx5xcos3xsin4
7xcosxsin7I
ta vit C3sinx)-B(4cosx)5cosx3A(4sinx7cosx7sinx ++++=+
Khi A = 1; B = -1; C = 2 v +++++++
=
2/
0
2/
0
2/
0
dx5xcos3xsin4
2dx
5xcos3xsin4
)5xcos3xsin4(ddxI
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Xt ++=2/
0
1 dx5xcos3xsin4
2I
t2
xtant =
2t1
t2xsin
+=
2
2
t1
t1xcos
+
= suy ra
3
1dt
)2t(
12I
1
0
21=
+= . Vy ( ) 8
9ln
3
1
2I5xcos3xsin4lnxI 1
2/
0+=+++=
V)Phng php dng tch phn lin ktV d 1. Tnh:
1. +=2
0xcosxsin
xdxsinI
ta xt thm tch phn th hai: +=2
0xcosxsin
xdxcosJ
Khi :2
JI
=+ (*).
Mt khc 0xcosxsin
)xcosx(sind
xcosxsin
dx)xcosx(sinJI
2
0
2
0
=+
+=
+
=
(**). GiI h (*) v (**) suy ra I = J =4
.
2. dxxcosxsin
xsinI
2
0
nn
n
n +
=
ta xt dxxcosxsin
xcosJ
2
0
nn
n
n +
=
. Khi :2
JI nn
=+ (*)
Mt khc nu t x =2
- t th n
2
0nn
n2
0nn
n
n Jdxxcosxsin
xcosdt
tcostsin
tcosI =
+
=
+
=
(**). T (*), (**) ta c4
In
=
3. dxxcosxsin
xsinI
2
0
nn
n
n +
=
tng t xt dxxcosxsin
xcosJ
2
0
nn
n
n +
=
v suy ra4
JI nn
==
4. dxxcos3xsin
xsinE
6
0
2
+
=
v dxxcos3xsin
xcosF
6
0
2
+
=
ta c 3ln4
1dx
xcos3xsin
1FE
6
0
=
+
=+
(*)
Li c 31dx)xcos3x(sinF3E6
0
==
(**). GiI h (*), (**) ta c:4
313ln
16
1E
= v
4
313ln
16
3F
+= . M rng tnh ==
+
= EFdxxcos3xsin
x2cosE
6
0
2
313ln
8
1 +
xut dxxcos3xsin
x2cosL
6
0
=
Cc bi ton tng t.Cc bi ton tng t.Cc bi ton tng t.Cc bi ton tng t.A Phng php bin i trc tip
1. [HNNI.98.A] ++++++++
====
1
0
x2
2x
e1
dx)e1(M
+ Bnh phng v phn tch thnh 2 phn s n gin.+ Bit i bin.
Gii: ++++++++++++++++
====
1
0
x2
x1
0
x2
x2
e1
dxe2
e1
dxe1M ta tnh ++++====
1
0
x2
x
1e1
dxe2M t (((( ))))2/;2/t,ttane x ==== khi vi tan =e v
+
=
4/
221tcos)ttan1(
tdttan2M =
2
e1ln
ttan1
1ln2tcosln2tdttan2
2
4/
24/
4/
+=
+==
2. [HTCKT.97] +2
0
3
xcos1xdxsin3
+
Gii:
2. [HTCKT.97] +2
0
3
xcos1xdxsin3
+
Gii:
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2. [HTCKT.97] +2
0
3
xcos1xdxsin3
+
Gii:
2. [HTCKT.97] +2
0
3
xcos1xdxsin3
+
Gii:
1. [HNNI.98.A] +
+1
0
x2
2x
e1
dx)e1(
2. [HTCKT.97] +2
0
3
xcos1xdxsin3
3. [HBK.98]
+
2
0
44 dx)xcosx(sinx2cos
4. [HDL.98] ++
2
11x1x
dx
5. +
6
0
dx)
6xcos(.xcos
1
6. +
2e
e
dxx
)xln(lnxln
7. [HM.00] +
3
6dx
)6
xsin(xsin
1
8. 3
0
4 xdx2sinxcos
9. [HNN.01] +4
0
66dx
xcosxsin
x4sin
10. [HNNI.01] 2
4
4
6
dxxsinxcos
11.
3
4
4xdxtg
12. [CGTVT.01]
+
3
2
2 dx.x3x
13. [CSPBN.00] +3
0
2 dx4x4x
14.
0
dxxsinxcos
15. +3
6
22 dx2xgcotxtg
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16. +3
0
23 dxxx2x
17.
1
1
dxx4 p: )35(2
18.
1
1
dxxx 3
22
19. ( )
+
5
3
dx2x2x 8
20. +
+3
0
2
2
dx.2xx
1xx
21. +
0
dxx2cos22 4
22.
0dxx2sin1 22
23. +2/
0
dxxsin1
24
24. 1
0
Ra;dxax
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B Phng php i bin
1. [CBN.01] +1
0
32
3
dx)x1(
x HD t fsf
2. [PVB.01] 1
0
23 dx.x1x
3. +
3ln
0
x 2e
dx
4. [CXD.01] +2
0
2dx
xcos1x2sin
5. [HKTQD.97] 1
0
635 dx)x1(x xut: 1
0
72 dx)x1(x
6. [HQG.97.B] +
1
0 x1
dx
7. [H.2004.A] +
2
1 1x1
xdx
``8. [H.2003.A] +
32
52 4xx
dx
9. [HSPHN.00.B]
0
222 dxxax
10. [HBK.00] +
2ln
0x
x2
1e
dxe
11. ++23
14 2x58xdx
12.( ) +
2
0
2dx
xsin2
x2sin
13. 4
0
3 xcos
dx
14. ++
6
2
dx1x4x2
1
15. +3
0
25 dxx1x
16. +
2e
e
dxx
)xln(lnxln
17. +
4
2
dxx
1x
18. ++
+++1
022
23
dx1x)x1(
x101x3x10
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19.
e
12
dxxln1x
1
20.
+
e
1
dxxln1x
xln
21. +++
4
03
dx1x21x2
1
23. +
4
72 9x.x
dx
24. ++
7
2
dx.2xx
1
25.( ) +
1
0
22x31
dx
26. [GTVT.00]
++++
2
22 dxxsin4xcosx
27. [HAN.97] ++++
0
2 xcos1
xdxsinx
28. [HLN.00] ++++++++2
0
dxxcosxsin2
1
29. [HH.00] ++++4
0
dxtgx1
1
30. [HVH.01] ++++
4
0 dxx2cosx2sin
xcosxsin
31. [HVBCVT.98] ++++2
0
2
3
xcos1
xdxcosxsin
32.
++++====
1
122
dx)1x(
1I
33. [HTN.01] ++++
++++
++++2)51(
124
2
dx1xx
1x
34. [HTCKT.00] ++++++++1
0
24dx
1xx
x
35. [HVKTQS.98]
++++++++++++
1
12 )x1x1(
dx
36. [PVBo.01] 1
0
23 dx.x1x
37. [H.2004.B] ++++
e
1
dxx
xlnxln31
38. [H.2005.A] ++++
++++2
0
dxxcos31
xsinx2sin
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39. [H.2006.A] ++++
2
022
dxxsin4xcos
x2sin
40. [H.2005.B] ++++2
0
dxxcos1
xcosx2sin
41. . [H.2005.B] ++++
5ln
3lnxx 3e2e
dx
42. [H.2003.A] ++++
32
52 4xx
dx
43. [H.2004.A] ++++
2
1
dx1x1
x
44. [H.2008.A] 6/
0
4
dxx2cos
xtan
45. [ thi th H] ++++4/
0
66dx
xcosxsinx4sin
46. [ thi th H]
++++
++++
e
1
2 dx.xln.xxln1.x
1
47. [ thi th H] ++++
4
0
1x2 dxe
48. [ thi th H] ++++2
0
3dx
)xsin1(2
x2sin
HD: t xsin1t ++++==== 81
t2
dt)1t(22
1
3====
49.
====
8
4
2
dxx 16xI
50. ++++++++
====
4
2
dxx
1x1xJ
51. ++++++++
++++++++++++====
1
022
23
dx1x)x1(
x101x3x10K
52.
====
2ln
2lnx2
x
dxe1
eH
53.
++++====
3ln
0x2 1e
dxG
54. ++++++++====2
0xcos3xsin53
dxF
55. ++++====2
0 24
3
dx3xcos3xcos
xcosD
56. ++++
====
5ln
0
x
xx
3e
dx1eeS
57. ++++
====
2 xcosxsin2
dxT
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58. ++++
====
4
72 9x.x
dxR
59. ++++++++
====
7
2
dx.2xx
1E
60. ++++++++
++++====
4
0
dx.1x21
1x2W
61. ++++====2/
0xcos2
dxQ
t2
xtant ==== th
9
3
t)3(
dtQ
utan3t1
022
====================
++++
====
62. ++++++++
====
2
02 1x6x3
dxM
C Phng php tch phn tng phn
1. [HC.97] ++++
1
0
x22 dxe)x1(
2. [HTCKT.98] 4
0
2 dx)1xcos2(x
3. ++++2
0
23 dx)1xln(x v 10
0
2 xdxlgx
4. [PVBo.98] e
1
2 dx)xlnx(
5. [HVNH.98]
0
2 xdxcosxsinx
6. [HC.00] ++++
2
12x
dx)1xln(
7. [HTL.01] ++++4
0
dx)tgx1ln(
8. 2
0
2xdxxtg
9. [HYHN.01] 3
2
2 dx.1x
10. [ thi th] 2
1
2 dx)xx3ln(x
11. [H.2007.D] e
1
22 xdxlnx
12. [H.2006.D] 1
0
x2 dxe)2x(
13.
++++++++====
0
1
3x2 dx)1xe(xI
14. ++++====
2e
e
dxx
)xln(lnxlnJ
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15. ====
0
2 dx)xsinx(K
16. ++++====1
0
2dx
)1x2(sin
xH
17. ====4
0
3dx
xcos
xsin.xG
18. ====2ln
0
x5 dxe.xF2
19. ++++
++++====
4
1
dxxx
)1xln(D
20.
++++
++++
====
e
1
2 dxxlnxln1x
xlnS
21. ====2
2 4/
2 dx.xcosA
22. (((( ))))====
0
2x dxxcos.eP
23. dx.xsin.xU
2
0====
24. dx.xcos.xsin.xY 2
0====
25. ++++++++
====
3/
0
xdxexcosxsin
xsin1T
t ====++++
++++==== dv;xcosxsin xsin1u ..Xt ++++====
3/
0
x1 dxexcos1 xsinT
v t tptp suy ra
3
e
xcos1
xsineT
33/
0
x
====++++
====
26. ++++====1
0
2 dxx1R s:2
)21ln(2 ++++++++
27. ++++====1
0
2 dx)1xln(xE S:2
12ln
28.
++++====
2/
0
dx)xcos1ln(xcosW
s: 12
29. ++++====e
e/12
dx)1x(
xlnQ s:
1e
e2++++
30. ++++++++
====
2/
3/
dxxcos1xsinx
M
Vit M = M1+ M2. Vi
23
lndxxcos1
xsinM
2/
3/
1 ====++++
====
& ++++====3/
6/
2 dxxcos1
xM
.
t
++++====
====
dxxcos1
1dv
xu
====
====
2
xcot2v
dxdu 4ln
3
)323(M 2
====
Vy
8
3ln
3
)323(M ++++
====
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31. ++++++++
====
2/
6/
dxx2cos1x2sinx
M
B Phng php h s bt nh
1. [HYHN.00] ++++2
12
2
12x7xdxx
2. [HNNI.00] ++++2
1
3dx
)1x(x
1v ++++
1
0
3dx
1x
3
3. [HXD.98] ++++++++
4
0
dxxsin3xcos4
xsin2xcos
4. [HTM.00] ++++2
0
3dx
)xcosx(sin
xsin4
5. [HTN.98]
++++++++
1
0 n nn x1)x1(
dx
6. ++++++++
====
2
0
2
4
dx4x
1xxI
7. ++++++++++++
====
1
03
2
dx1x
7x3x2J
8. ++++++++++++
====
1
0
2dx
2x3x
5x4K
9.(((( ))))
====
1
0
22 4x3x
dxL
10. ++++++++
====
2
0
2
4
dx.4x
1xxZ
D Phng php tch phn lin kt
1. ++++
2
0
dxxcosxsin
xcos
2. [ thi th] ++++
++++
====
32
32
x
1x
3
4
dxex
1xI
2
HD: )()1
( xFx
F = . Suy ra
0)32
1()32( =+
+= FFI
[HTN.00] CMR: Zn , ta c
0dx)nxxsin(sin2
0
====++++
[HVKTQS.01] ++++
b
0
22
2
dx)xa(
xa, 0b,a >>>>
[HLN.01] ++++++++
1
0
2
x2
dx)1x(
e)1x(