lista de exercícios 32 - integração por partes -...

UNEMAT – Universidade do Estado de Mato Grosso Campus Universitário de Sinop

Faculdade de Ciências Exatas e Tecnológicas Curso de Engenharia Civil

Disciplina: Cálculo Diferencial e Integral I

Página 1 de 22

Lista de Exercícios – Integração por Partes

1) Nos exercícios abaixo, calcule a integral indefi nida por integração

por partes. a) 3xxe dx∫

3xdv e dx=

v dv= ∫ u x=

3xv e dx= ∫ du dx=

313

xv e=

u dv uv v du= −∫ ∫

3 3 31 13 3

x x xxe dx x e e dx= ⋅ −∫ ∫

3 3 31 13 3


3 3 31 1 13 3 3

x x xxe dx x e e C= ⋅ − ⋅ ⋅ +∫

3 3 31 13 9

x x xxe dx xe e C= − +∫

33 1

3 3

xx e

xe dx x C = − +

∫

b) 2 xx e dx−∫

v dv= ∫ 2u x=

xv e dx−= ∫ 2du x dx= xv e−= −


2 2 2x x xx e dx x e e xdx− − −= − − −∫ ∫ 2 2 2x x xx e dx x e xe dx− − −= − +∫ ∫




Página 2 de 22

xxe dx−∫

v dv= ∫ u x=

xv e dx−= ∫ du dx= xv e−= −


x x xxe dx xe e dx− − −= − − −∫ ∫ x x xxe dx xe e dx− − −= − +∫ ∫ x x xxe dx xe e− − −= − −∫

Portanto:

2 2 2x x xx e dx x e xe dx− − −= − +∫ ∫

( )2 2 2x x x xx e dx x e xe e C− − − −= − + − − +∫ 2 2 2 2x x x xx e dx x e xe e C− − − −= − − − +∫

( )2 2 2 2x xx e dx e x x C− −= − + + +∫

c) ( )ln 2x dx∫

v dv= ∫ ln(2 )u x=

v dx= ∫ 1

22

du dxx

= ⋅

v x= 1

du dxx

=


1ln(2 ) ln(2 )x dx x x x dx

x= ⋅ − ⋅∫ ∫

ln(2 ) ln(2 )x dx x x dx= ⋅ −∫ ∫

ln(2 ) ln(2 )x dx x x x C= ⋅ − +∫

[ ]ln(2 ) ln(2 ) 1x dx x x C= − +∫

2) Nos exercícios abaixo, calcule a integral indefi nida (Sugestão: a

integração por partes pode não ser necessária para todas as integrais).




Página 3 de 22

a) 4xe dx∫

4 4u x du dx= ⇒ =

4 4 41 1 1 1

44 4 4 4

x x u u xe dx e dx e du e C e C= = = + = +∫ ∫ ∫

b) 4xxe dx∫

v dv= ∫ u x=


414

xv e=


4 4 41 14 4


4 4 41 14 4

x x xxe dx xe e dx= −∫ ∫

4 4 41 1 14 4 4

x x xxe dx xe e C= − ⋅ +∫

4 4 41 14 16

x x xxe dx xe e C= − +∫

4 41 14 4

x xxe dx e x C = − +

∫

c)

2xxe dx∫

2 2 2

2 2x x xu e du e x dx du xe dx= ⇒ = ⋅ ⇒ =

2 2 21 1 1 12

2 2 2 2x x u u xxe dx xe dx e du e C e C= = = + = +∫ ∫ ∫

d) 3 xx e dx∫

v dv= ∫ 3u x=

xv e dx= ∫ 23du x dx= xv e=




Página 4 de 22


( )3 3 23x x xx e dx x e e x dx= −∫ ∫ 3 3 23x x xx e dx x e x e dx= −∫ ∫

2 xx e dx∫

v dv= ∫ 2u x=

xv e dx= ∫ 2du xdx= xv e=


( )2 2 2x x xx e dx x e e x dx= −∫ ∫ 2 2 2x x xx e dx x e xe dx= −∫ ∫

xxe dx∫

v dv= ∫ u x=

xv e dx= ∫ du dx= xv e=


x x xxe dx xe e dx= −∫ ∫ x x xxe dx xe e= −∫

Portanto:

3 3 23x x xx e dx x e x e dx= −∫ ∫

( )3 3 23 2x x x xx e dx x e x e xe dx= − −∫ ∫

3 3 23 6x x x xx e dx x e x e xe dx= − +∫ ∫

( )3 3 23 6x x x x xx e dx x e x e xe e C= − + − +∫ 3 3 23 6 6x x x x xx e dx x e x e xe e C= − + − +∫

( )3 3 23 6 6x xx e dx e x x x C= − + − +∫




Página 5 de 22

e) 3 lnx x dx∫

v dv= ∫ lnu x=

3v x dx= ∫ 1

du dxx

=

414

v x=


3 4 41 1 1ln ln

4 4x x dx x x x dx

x= − ⋅∫ ∫

3 4 31 1ln ln

4 4x x dx x x x dx= −∫ ∫

43 41 1ln ln

4 4 4x

x x dx x x C= − ⋅ +∫

3 4 41 1ln ln

4 16x x dx x x x C= − +∫

3 41 1ln ln

4 4x x dx x x C = − +

∫

f) ( )ln 1t t dt+∫

v dv= ∫ ( )ln 1u t= +

v t dt= ∫ 1

1du dt

t=

+

212

v t=


( ) ( )2 21 1 1ln 1 ln 1

2 2 1t t dt t t t dt

t+ = ⋅ + − ⋅

+∫ ∫

( ) ( )2 21

ln 1 ln 12 2 1t t

t t dt t dtt

+ = + −+∫ ∫

( )2 21

1 1 1 2 1

t t tt t t tdt dt t dt dt dt

t t t t

+ −= = − = −

+ + + +∫ ∫ ∫ ∫ ∫




Página 6 de 22

1t

dtt +∫

1 1u t du dt t u= + ⇒ = ⇒ = −

( )1 1ln ln 1

1t u

dt dt dt dt t u t tt u u

−= = − = − = − ++∫ ∫ ∫ ∫

Portanto:

( ) ( )2 21

ln 1 ln 12 2 1t t

t t dt t dtt

+ = + −+∫ ∫

( ) ( )2 21

ln 1 ln 12 2 2 1t t t

t t dt t dtt

+ = + − − +

∫ ∫

( ) ( )2 2 1

ln 1 ln 12 4 2 1t t t

t t dt t dtt

+ = + − ++∫ ∫

( ) ( ) ( )2 2 1

ln 1 ln 1 ln 12 4 2t t

t t dt t t t C + = + − + − + + ∫

( ) ( ) ( )2 2 1

ln 1 ln 1 ln 12 4 2 2t t t

t t dt t t C+ = + − + − + +∫

( ) ( ) ( )2 21 2

ln 1 ln 1 ln 12 2 4 4t t t

t t dt t t C+ = + − + − + +∫

( ) ( ) ( ) ( )2 1 2ln 1 ln 1

2 4

t t tt t dt t C

− −+ = + − +∫

g) ( )2

lnx x dx∫

v dv= ∫ ( )2

lnu x=

v x dx= ∫ 1

2lndu x dxx

= ⋅

212

v x=


( ) ( )2

2 2 21 1ln ln 2ln

2 2x

x x dx x x x dxx

= ⋅ − ⋅ ⋅∫ ∫

( ) ( )2

2 2ln ln ln

2x

x x dx x x x dx= ⋅ −∫ ∫




Página 7 de 22

lnx x dx∫

v dv= ∫ lnu x=

v x dx= ∫ 1

du dxx

=

212

v x=


221 1

ln ln2 2x

x x dx x x dxx

= ⋅ − ⋅∫ ∫

2 1ln ln

2 2x

x x dx x x dx= ⋅ −∫ ∫

2 21ln ln

2 2 2x x

x x dx x C= ⋅ − ⋅ +∫

2 2

ln ln2 4x x

x x dx x C= ⋅ − +∫

Portanto:

( ) ( )2

2 2ln ln ln

2x

x x dx x x x dx= ⋅ −∫ ∫

( ) ( )2 2 2

2 2ln ln ln

2 2 4x x x

x x dx x x C

= ⋅ − ⋅ − +

∫

( ) ( )2 2 2

2 2ln ln ln

2 2 4x x x

x x dx x x C= ⋅ − ⋅ + +∫

h) ( )2ln x

dxx∫

lnu x=

1

du dxx

=

( ) ( ) ( )2 332 2ln ln1

ln3 3

x xudx x dx u du C C

x x= = = + = +∫ ∫ ∫




Página 8 de 22

i) 1x x dx−∫

1 1u x du dx x u= − ⇒ = ⇒ = +

( ) ( ) ( )31 12 2 21 1 1x x dx u u du u u du u u du− = + ⇒ + = + =∫ ∫ ∫ ∫

( )5 3

2 23 5 312 2 2 22 2

5 3 5 32 2

u uu u du C u u C= + = + + = + + =∫

( ) ( )5 32 22 2

1 15 3

x x C= − + − +

j) ( )2 1 xx e dx−∫

v dv= ∫ 2 1u x= −



( ) ( )2 21 1 2x x xx e dx e x e x dx− = − − ⋅∫ ∫

( ) ( )2 21 1 2x x xx e dx e x xe dx− = − −∫ ∫

xxe dx∫

v dv= ∫ u x=




Portanto:

( ) ( )2 21 1 2x x xx e dx e x xe dx− = − −∫ ∫

( ) ( ) ( )2 21 1 2x x x xx e dx e x xe e C− = − − − +∫

( )2 21 2 2x x x x xx e dx e x e xe e C− = − − + +∫




Página 9 de 22

( )2 21 2x x x xx e dx e x e xe C− = + − +∫

( ) ( )2 21 2 1x xx e dx e x x C− = − + +∫

( ) ( )22 1 1x xx e dx e x C− = − +∫

k) ( )

2

22 1

xxedx

x +∫

2 22x xu e du e dx= ⇒ =

v dv= ∫

( )22 1

xv dx

x=

+∫

1

2 1 2 12

ww x x w x

−= + ⇒ = − ⇒ =

22

dwdw dx dx= ⇒ =

2

1 12 2

w dwv

w−= ⋅ ⋅∫

2

14w

v dww−= ∫

2

1 14

wv dw

w−= ∫

2

1 1 14

v dw dww w

= − ∫ ∫

21 14

v dw w dww

− = − ∫ ∫

11ln

4 1w

v w−

= − −

1 1ln

4v w

w = +

( )1 1ln 2 1

4 2 1v x

x = + + +

Portanto:





Página 10 de 22

( )( ) ( )

22 2

2

1 1 1 1ln 2 1 ln 2 1 2

4 2 1 4 2 12 1

xx xxe

dx e x x e dxx xx

= ⋅ + + − + + ⋅ + + +∫ ∫

( )( ) ( )

2 22

2

1 1 1ln 2 1 ln 2 1

4 2 1 2 2 12 1

x xxxe e

dx x x e dxx xx

= + + − + + + + +∫ ∫

( ) ( )2

2 21ln 2 1 ln 2 1

2 1 2 1

xx x e

x e dx e x dx dxx x

+ + = + + + + ∫ ∫ ∫

( )2 ln 2 1xe x dx + ∫

v dv= ∫ ln(2 1)u x= +

2xv e dx= ∫ 1

22 1

du dxx

= ⋅+

212

xv e= 2

2 1du dx

x=

+


( )2 2 21 1 2ln 2 1 ln(2 1)

2 2 2 1x x xe x dx e x e dx

x + = ⋅ + − ⋅ +∫ ∫

( )2

2 21ln 2 1 ln(2 1)

2 2 1

xx x e

e x dx e x dxx

+ = ⋅ + − +∫ ∫

Portanto:

( )( ) ( )

2 2 22

2

1 1ln 2 1 ln 2 1

4 2 1 2 2 12 1

x x xxxe e e

dx x e x dx dxx xx

= + + − + + + + + ∫ ∫ ∫

( )( ) ( )

2 2 22

2

1 1 1ln 2 1 ln 2 1

4 2 1 2 2 2 12 1

x x xxxe e e

dx x e x dx dxx xx

= + + − + − + + +∫ ∫ ∫

( )( )

2 2 2 22

2

1 1 1 1ln 2 1 ln(2 1)

4 2 1 2 2 2 1 2 2 12 1

x x x xxxe e e e

dx x e x dx dxx x xx

= + + − ⋅ + − − + + + + ∫ ∫ ∫

( )( )

2 22

2

1 1ln 2 1 ln(2 1)

4 2 1 42 1

x xxxe e

dx x e x Cxx

= + + − ⋅ + + + +∫

( ) ( )2 2

2 4 2 12 1

x xxe edx C

xx= +

++∫




Página 11 de 22

3) Calcule a integral definida.

a) 1

2

0

xx e dx∫

v dv= ∫ 2u x=



2 2 2x x xx e dx e x e x dx= − ⋅∫ ∫ 2 2 2x x xx e dx e x xe dx= −∫ ∫

xxe dx∫

v dv= ∫ u x=




Portanto:

2 2 2x x xx e dx e x xe dx= −∫ ∫

( )2 2 2x x x xx e dx e x xe e C= − − +∫ 2 2 2 2x x x xx e dx e x xe e C= − + +∫

( )2 2 2 2x xx e dx e x x C= − + +∫

( )1 1

2 2

00

2 2x xx e dx e x x = − + ∫

( ) ( )1

2 1 2 0 2

0

1 2 1 2 0 2 0 2xx e dx e e = − ⋅ + − − ⋅ + ∫




Página 12 de 22

12

0

2xx e dx e= −∫

b) 5

1

lne

x x dx∫

v dv= ∫ lnu x=

5v x dx= ∫ 1

du dxx

=

616

v x=


65 61 1ln ln

6 6x

x x dx x x dxx

= − ⋅∫ ∫

65 51ln ln

6 6x

x x dx x x dx= −∫ ∫

6 65 1ln ln

6 6 6x x

x x dx x C= − ⋅ +∫

65 1ln ln

6 6x

x x dx x C = − +

∫

6

5

1 1

1ln ln

6 6

ee xx x dx x

= −

∫

6 65

1

1 1 1ln ln ln1

6 6 6 6

e ex x dx e

= − − −

∫

65

1

1 1 1ln 1

6 6 6 6

e ex x dx

= − − −

∫

65

1

5 1ln

6 6 36

e ex x dx = ⋅ +∫

65

1

5 1ln

36 36

e ex x dx = +∫

( )5 6

1

1ln 5 1

36

e

x x dx e= +∫




Página 13 de 22

4) Nos exercícios abaixo, calcule a integral indefi nida, utilizando o método indicado. a) 2 2 3x x dx−∫

a.1) Por partes, fazendo 2 3dv x dx= − . v dv= ∫ 2u x=

2 3v x dx= −∫ 2du dx=

2 3w x= −

2dw dx=

12 2 3

2v x dx= −∫

12

v w dw= ∫

121

2v w dw= ∫

321

322

wv = ⋅

321 2

2 3v w= ⋅

321

3v w=

( )321

2 33

v x= −


( ) ( )3 32 21 1

2 2 3 2 2 3 2 3 23 3

x x dx x x x dx− = ⋅ − − − ⋅∫ ∫

( ) ( )3 32 22 2

2 2 3 2 3 2 33 3x

x x dx x x dx− = − − −∫ ∫

( )322 3x dx−∫

2 3u x= −

2du dx=




Página 14 de 22

( ) ( )5

23 3 3 52 2 2 21 1 1 1 2

2 3 2 2 352 2 2 2 5

2

ux dx x dx u du u− = − = = ⋅ = ⋅ =∫ ∫ ∫

( )55221 1

2 35 5

u x= = −

Portanto:

( ) ( )3 32 22 2

2 2 3 2 3 2 33 3x

x x dx x x dx− = − − −∫ ∫

( ) ( )3 52 22 2 1

2 2 3 2 3 2 33 3 5x

x x dx x x C − = − − − + ∫

( ) ( )3 52 22 2

2 2 3 2 3 2 33 15x

x x dx x x C− = − − − +∫

a.2) Por substituição, fazendo 2 3u x= − .

2 3u x= − 2 3u x= − 2 2 3u x= − ( )1

22 3u x= −

22 3x u= + ( ) 121

2 3 22

du x dx−= − ⋅

( )12

1

2 3du dx

x=

−

1

2 3du dx

x=

−

2 3dx x du= − dx u du=

( ) ( )2 4 2 5 31 12 2 3 3 3 3

5 3x x dx u u udu u u du u u C− = + ⋅ = + = + ⋅ + =∫ ∫ ∫

( ) ( ) ( ) ( )5 3 5 35 3 2 21 1 1

2 3 2 3 2 3 2 35 5 5

u u C x x C x x C= + + = − + − + = − + − +

OBS: Diferencie as duas expressões obtidas para comprovar que ambas têm a mesma integral.




Página 15 de 22

b) 4 5

xdx

x+∫

b.1) Por partes, fazendo 1

4 5dv dx

x=

+.

v dv= ∫ u x=

1

4 5v dx

x=

+∫ du dx=

4 5w x= +

5dw dx=

1

4 5v dx

x=

+∫

1 55 4 5

v dxx

=+∫

1 15

v dww

= ∫

121

5v w dw

−= ∫

121

152

wv = ⋅

121

25

v w= ⋅ ⋅

122

5v w=

( ) 122

4 55

v x= +


( ) ( )1 12 22 2

4 5 4 55 54 5

xdx x x x dx

x= + ⋅ − +

+∫ ∫

( ) ( )1 12 22 2

4 5 4 55 54 5

x xdx x x dx

x= + − +

+∫ ∫




Página 16 de 22

( ) 124 5x dx+∫

4 5u x= + 5u dx=

( ) ( )3

2 31 1 12 2 2 21 1 1 1 2

4 5 5 4 535 5 5 5 3

2

ux dx x dx u du u+ = + = = ⋅ = ⋅ =∫ ∫ ∫

( )33222 2

4 515 15

u x= = +

Portanto:

( ) ( )1 12 22 2

4 5 4 55 54 5

x xdx x x dx

x= + − +

+∫ ∫

( ) ( )312 22 2 2

4 5 4 55 5 154 5

x xdx x x C

x = + − + + +

∫

( ) ( )312 22 4

4 5 4 55 754 5

x xdx x x C

x= + − + +

+∫

b.2) Por substituição, fazendo 4 5u x= + .

4 5u x= + 4 5u x= + 2 4 5u x= + ( )1

24 5u x= +

25 4x u= − ( ) 121

4 5 52

du x dx−= + ⋅

2 45

ux

−= ( ) 125

4 52

du x dx−= +

( )1

2

5

2 4 5du dx

x=

+

5

2 4 5du dx

x=

+

5

2du dx

u=

25

dx u du=




Página 17 de 22

( )2

2 34 1 2 2 2 14 4

5 5 25 25 34 5

x udx u du u du u u C

ux

− = ⋅ ⋅ = − = − + = + ∫ ∫ ∫

( ) ( )332 8 2 8

4 5 4 575 25 75 25

u u C x x C= − + = + − + + =

( ) ( )3 13 2 22 8 2 84 5 4 5

75 25 75 25u u C x x C= − + = + − + +

OBS: Diferencie as duas expressões obtidas para comprovar que ambas têm a mesma integral.

5) Utilize a integração por partes para verificar a s fórmulas abaixo:

( )( )

1

2ln 1 1 ln 11

nn x

x x dx n x C nn

+

= − + + + ≠ − +∫

v dv= ∫

lnu x=

nv x dx= ∫ 1

du dxx

=

1

1

nxv

n

+

=+

u dv uv v du= −∫ ∫ 1 1 1

ln ln1 1

n nn x x

x x dx x dxn n x

+ +

= ⋅ − ⋅+ +∫ ∫

1 1ln ln

1 1

nn nx

x x dx x x dxn n

+

= ⋅ −+ +∫ ∫

1 11ln ln

1 1 1

n nn x x

x x dx xn n n

+ +

= ⋅ − ⋅+ + +∫

1 1ln ln

1 1

nn x

x x dx xn n

+ = − + + ∫

( )1 1 ln 1ln

1 1

nn n xx

x x dxn n

+ + −= + +

∫

( )( )

1

2ln 1 1 ln1

nn x

x x dx n xn

+

= − + + +∫




Página 18 de 22

1n ax

n ax n axx e nx e dx x e dx

a a−= −∫ ∫

v dv= ∫

nu x=

axv e dx= ∫ 1ndu nx dx−=

1 axv ea

=


11 1n ax n ax ax nx e dx x e e nx dxa a

−= ⋅ − ⋅∫ ∫

1n ax

n ax n axx e nx e dx x e dx

a a−= −∫ ∫

6) Nos exercícios a seguir, calcule a integral inde finida com auxílio

das fórmulas acima. a) 2 5xx e dx∫

1 2 e 5n ax

n ax n axx e nx e dx x e dx n a

a a−= − = =∫ ∫

2 5 2 5 51 25 5

x x xx e dx x e xe dx= −∫ ∫

5xxe dx∫

v dv= ∫

u x=


515

xv e=


5 5 51 15 5


5 5 515 5

x x xxxe dx e e dx= −∫ ∫

5 5 51 15 5 5

x x xxxe dx e e= − ⋅∫




Página 19 de 22

5 5 515 25

x x xxxe dx e e= −∫

Portanto:

2 5 2 5 51 25 5

x x xx e dx x e xe dx= −∫ ∫

2 5 2 5 5 51 2 15 5 5 25

x x x xxx e dx x e e e C = − − +

∫

2 5 2 5 5 51 2 25 25 125

x x x xx e dx x e xe e C= − + +∫

( )5

2 5 225 10 2125

xx e

x e dx x x C= − + +∫

b) 2 lnx x dx−∫

( )( )

1

2ln 1 1 ln 21

nn x

x x dx n x nn

+

= − + + = − +∫

( )( )

2 12

2ln 1 2 1 ln2 1

xx x dx x C

− +− = − + − + + − +∫

( )2 1ln 1 lnx x dx x x C− −= − − +∫

( )2 1ln ln 1x x dx x C

x− = − + +∫

7) Determine a área da região delimitada pelos gráf icos das equações

dadas.

, 0, 4xy xe y x−= = =

0,00

0,05

0,10

0,15

0,20

0,25

0,30

0,35

0,40

0,00 1,00 2,00 3,00 4,00 5,00




Página 20 de 22

4

0

xA xe dx−= ∫

1 1 e 1n ax

n ax n axx e nx e dx x e dx n a

a a−= − = = −∫ ∫

1 1

1 1 111 1

xx xx e

xe dx x e dx−

− − −= −− −∫ ∫

x xx

xxe dx e dx

e− −= − +∫ ∫

x xx

xxe dx e

e− −= − −∫

1xx x

xxe dx

e e− = − −∫

( )11x

xxe dx xe

− = − +∫

( )44

00

11x

xA xe dx xe

− = = − +

∫

( ) ( )4

4 00

1 14 1 0 1xA xe dx

e e− = = − + − − +

∫

4

40

51xA xe dx

e−= = − +∫

44

0

1 5xA xe dx e− −= = −∫

8) Dada a região delimitada pelos gráficos de 2lny x= , 0y = e x e= ,

determine:

a) A área da região.

0,00

0,50

1,00

1,50

2,00

2,50

0,00 0,50 1,00 1,50 2,00 2,50 3,00




Página 21 de 22

0 2ln 0 ln 0 1y x x x= ⇒ = ⇒ = ⇒ =

1 1

2ln 2 lne e

A x dx x dx= =∫ ∫

ln x dx∫

v dv= ∫

lnu x=

v dx= ∫ 1

du dxx

=

v x=

u dv uv v du= −∫ ∫ 1

ln lnx dx x x x dxx

= − ⋅∫ ∫

ln lnx dx x x dx= −∫ ∫

ln lnx dx x x x= −∫

( )ln ln 1x dx x x= −∫

( ) ( ) ( ){ }11

2 ln 2 ln 1 2 ln 1 1 ln1 1e

eA x dx x x e e = = − = − − − ∫

( ) ( ){ }1

2 ln 2 1 1 1 0 1 2e

A x dx e = = − − − = ∫

b) O volume do sólido gerado pela revolução da regi ão em torno

do eixo x .

( ) ( )2 2

1 1

2ln 4 lne e

V x dx x dx= =∫ ∫π π

( )2ln x dx∫

v dv= ∫ ( )2

lnu x=

v dx= ∫ 1

2lndu x dxx

=

v x=




Página 22 de 22


( ) ( )2 2 1ln ln 2lnx dx x x x x dx

x= − ⋅∫ ∫

( ) ( )2 2ln ln 2 lnx dx x x x dx= −∫ ∫

( ) ( ) ( )2 2ln ln 2 lnx dx x x x x x= − −∫

( ) ( )2 2ln ln 2 ln 2x dx x x x x x= − +∫

( ) ( )2 2

11

4 ln 4 ln 2 ln 2e e

V x dx x x x x x = = − + ∫π π

( ) ( ){ }2 24 ln 2 ln 2 1 ln1 2 1 ln1 2 1V e e e e e = − + − − ⋅ ⋅ + ⋅

π

( )4 2 2 2V e e e = − + − π

( )4 2V e= −π

lista de exercícios 32 - integração por partes -...

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