Post on 15-Jan-2016
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INTEGRACION POR FRACCIONES PARCIALES(LAB1)
P2.
3)
∫ aex−b
a ex+b
∫ aex−b
a ex+bdx=∫(1+ 2b
aex−b )dx=∫ dx+∫ 2baex−b
dx=x+C1+2∫ bdxa ex−b
∫ aex−b
a ex+b=x+2∫ bdx . e− x
(a ex−b ) . e− x+C1= x+2∫ be−x dx
a−be−x+C1
u=a−be−xdu=d (a−be− x)du=(a−be−x )' dxdu=−1−1be− xdxdu=be− x
∫ aex−b
aex−b=x+2∫ du
u+C1=x+2 lnu+C1+C2
∫ aex−b
a ex+b=x+2 ln (a−be−x )+C
P3.
2)
∫ (cscθ)2dθ√2cotθ+3
u=2cot θ+3du=d (2cotθ+3 )du=(2cot θ+3 )' dθdu=2 (csc θ )2dθdu2
=(csc θ)2dθ
∫ ( cscθ )2dθ√2cotθ+3
=∫ du2.1√u
=12∫ (u )
−12 du=1
2(u )
12
12
+C=√u+C
∫ (cscθ)2dθ√2cotθ+3
=√2cot θ+3+C
P4.
4)
∫ dx
x [1+(ln x)2 ]
u=ln xdu=d ( ln x )du=( ln x )'dxdu=dxx
∫ dx
x [1+( ln x )2 ]=∫ du
1+u2=11tan−1( u1 )+C=tan−1u+C
∫ dx
x [1+(ln x)2 ]=tan−1( ln x )+C
P6.
1)
∫ ( tan x )4 (sec x )2dx
u=tan xdu=d ( tan x )du=( tan x )' dxdu=( sec x )2dx
∫ ( tan x )4 (sec x )2dx=∫u4du=u3
3+C
∫ ( tan x )4 (sec x )2dx= ( tan x )3
3+C
P7.
3)
∫ x ex
(1+x )2dx
Integramos por partes:
u=xexdu=d (x ex)du=(x ex )'dx
du=[x ' ex+x (ex )' ]dxdu=[ex+x ex ]dxdu=ex (1+ x )dx
dv= dx
(1+x )2∫ dv=∫ dx
(1+x )2v=∫ dx
(1+x )2
(1+x )=td (1+x )=dt (1+x )'dx=dtdx=dt
v=∫ dtt 2
= t−1
−1+C=−1
t+Cv=
−1(1+x )
+C
∫ x ex
(1+x )2dx=x ex [ −1
(1+x ) ]−∫ [ −1(1+x ) ]ex (1+x )dx=−x ex
(1+x )+∫ e
x (1+x )(1+x )
dx
∫ x ex
(1+x )2dx=∫e xdx− x ex
(1+x )=ex+C1−
x ex
(1+x )+C2=e
x− x ex
(1+x )+C
∫ x ex
(1+x )2dx=e x( 1−x1+x )+C
P7.
8)
∫ tan−1 (√ x+1 )dx
t=x+1dt=d ( x+1 )dt=( x+1 )' dxdt=dx
∫ tan−1(√t)dt
k=√ tdk=d (√ t )dk=(√ t )dtdk=dt2√ t
= dt2k
2kdk=dt
∫ tan−1 (√ x+1 )dx=∫ tan−1(√t )dt=∫ tan−1 k .2kdk=2∫ tan−1 k dk
Integramos por partes: 2∫ tan−1k dk
u=tan−1kdu=d (tan−1 k )du=( tan−1 k )' dk
du= dk
1+k2
dv=kdk∫ dv=∫kdk v= k2
2+C
2∫ tan−1k dk=2 tan−1k .k2
2−2∫ k
2
2dk
(1+k2 )=k2 . tan−1 k−∫ k2dk
(1+k2 )
2∫ tan−1k dk=k 2 . tan−1
k−∫(1− 1
1+k2 )dk=k2 . tan−1 k−∫ dk+∫ dk
1+k2
2∫ tan−1k dk=k 2 . tan−1k−k+C1+11tan−1( k1 )+C2=k2 . tan−1 k−k+ tan−1 k+C
k=√ t
∫ tan−1 (√ t )dt=2∫ tan−1 k dk=(√t )2. tan−1√t−√ t+ tan−1√ t+C
∫ tan−1 (√ t )dt=t . tan−1√t−√t+tan−1√ t+C
t=x+1
∫ tan−1 (√ x+1 )dx=∫ tan−1 (√ t )dt= (x+1 ) tan−1√ x+1−√ x+1+ tan−1√ x+1+C
∫ tan−1 (√ x+1 )dx= ( x+2 ) tan−1√x+1−√x+1+C
P8.
5)
∫ 2x3
(x2−1 )4dx
∫ 2x3
(x2−1 )4dx=2∫ x3dx
(x2−1 )4
Resolvemos por fracciones parciales:
x3
(x2−1 )4=Ax+Bx2−1
+Cx+D
(x2−1 )2+Ex+F
( x2−1 )3+Gx+H
(x2−1 )4=
( Ax+B ) (x2−1 )3+(Cx+D ) (x2−1 )2+(Ex+F ) (x2−1 )+Gx+H(x2−1)4
x3=A x7+B x6+(C−3 A ) x5+ (D−3B ) x4+(E−2C ) x3+ (3 A+3B−2D+F ) x2+(G−E+C−A ) x+(H−F+D−B)
Resolviendo el sistema de ecuaciones tenemos:
A=0 ;B=0 ;C=0 ; D=0 ; E=1 ;F=0;G=1; H=0
x3
(x2−1 )4= x
(x2−1)3+ x( x2−1)4
2∫ x3dx
(x2−1 )4=2∫ xdx
(x2−1)3+2∫ xdx
(x2−1)4
u=x2−1du=d (x2−1 )du=(x2−1 )' dxdu=2 xdxdu2
=xdx
2∫ x3dx
(x2−1 )4=2∫ du
21u3
+2∫ du21u4
=∫u−3du+∫ u−4du=u−2
−2+C1+
u−3
−3+C2
2∫ x3dx
(x2−1 )4=−12u2
− 13u3
+C=−3u−26u3
+C
u=x2−1
2∫ x3dx
(x2−1 )4=
−3 ( x2−1 )−26 (x2−1 )3
+C=−3 x2+3−2
6 (x2−1 )3+C
2∫ x3dx
(x2−1 )4= 1−3x3
6 (x2−1 )3
P12.
1)
∫√x2+2x−8dx
∫√x2+2x−8dx=∫√ ( x+1 )2−32dx
u=x+1du=d ( x+1 )du=( x+1 )'dxdu=dx
∫√x2+2x−8dx=∫√u2−9du=12u√u2−9−1
2.32 ln (u+√u2−9 )+C
u=x+1
∫√x2+2x−8dx= ( x+1 ) √ ( x+1 )2−92
+9 ln [ ( x+1 )+√( x+1 )2−9 ]
2+C
∫√x2+2x−8dx=(x+1)√ x2+2 x−82
+9 ln [ ( x+1 )+√ x2+2 x−8]
2+C
7)
∫ ( x−1 )2
x2+3x+4dx
∫ ( x−1 )2
x2+3x+4dx=∫(1− 5 x−5
x2+3 x+4 )dx=∫dx−5∫ x−1x2+3 x+4
∫ ( x−1 )2
x2+3x+4dx=x+C1−5∫
[ 12 (2x+3 )−12 ]
x2+3x+4dx=x+C1−5.
12∫ 2 x+3x2+3 x+4
dx+5. 12∫ dxx2+3x+4
∫ ( x−1 )2
x2+3x+4dx=x−5
2ln (x2+3 x+4 )+C1+C2+
52∫ dx
( x+ 32 )2
+(√72 )2
∫ ( x−1 )2
x2+3x+4dx=x−
5 ln ( x2+3 x+4 )2
+C3+52
1
2 (√7 )ln( x+
32
√72
)+C4
∫ ( x−1 )2
x2+3x+4dx=x−
5 ln ( x2+3 x+4 )2
+
5√7 ln( x+32
√72
)28
+C