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