3 Integrales por partes (completo)

(1)

Integración por partes

∫

u·dv=u·v−

∫

v·du Normas para la aplicación de la formula de integración por partes:

a) A una parte de la integral le debemos llamar u y al resto dv

b) Debemos escoger dv, una parte que sea fácilmente integrable puesto que ∫dv = v y necesitamos v para aplicar la formula.

c) La integral que resulta

∫

v·du debe ser más sencilla que la propuesta. d) En algunos casos se deberá aplicar partes más de una vez.

Casos más frecuentes de aplicación de la integración por partes:

a) Función polinómica (u) por función exponencial (dv) b) Función polinómica (u) por función trigonométrica. (dv)

c) Función polinómica (dv) por función inversa de trigonométrica (u) d) Función polinómica (dv) por función logarítmica (u).

e) Función logarítmica (u) dv = dx

f) Funciones inversas de los trigonométricos (u) dv = dx g) Funciones trigonométricas por funciones exponenciales.

a)

∫

P(x)·eax·dx:

    

  

= =

ax

ax _e

a 1 v ; dx · e dv

(x)·dx P' du ); x ( P u

b)

∫

P(x)·senax·dx:

    

  

− = =

= =

ax sen 2

1 v ax·dx; sen dv

(x)·dx P' du ); x ( P u

c)

∫

P(x)·cosax·dx:

    

  

= =

ax sen a 1 v ; dx · ax cos dv

(x)·dx P' du ); x ( P u

d)

∫

P(x)·Ln x·dx:

    

    

= =

∫

P(x)·dx v

; dx )· x ( P dv

dx · x 1 du x; Ln u

e)

∫

P(x)·arctg x·dx:

     

   

= =

+ = =

∫

P(x)·dx v

; dx )· x ( P dv

dx · x 1

1 du x; arctg

u ₂

f)

∫

P(x)·arcsen x·dx:

     

     

= =

∫

P(x)·dx v

; dx )· x ( P dv

dx · x -1

1 du x; arcsen u

(2)

Ejemplos 1.

∫

= −

∫

= −

∫

= − +

    

  

= =

= dx x·Ln x dx x·Ln x x C

x 1 x x Ln · x x v ; dx dv

dx · x 1 du Ln x; u x·dx Ln

Ejemplo2.

∫

=

+ − =

    

  

=

= +

= =

= ·dx

x 1

x 2 1 x arctg · x dx · x 1

1 · x x arctg · x x

v ; dx dv

dx · x 1

1 du x; arctg u x·dx

arctg 2 ₂ ₂

C x 1 Ln · 2 1 x arctg ·

x − + 2 +

Ejemplo 3.

= −

=    

   

= =

=

∫

xe ·dx

3 2 3

e · x 2x·dx · e · 3 1 e · 3 1 · x e · 3 1 v ; dx · e dv

2x·dx du ; x u dx e ·

x ₃_x ₃_x 2 3x 3x 2 3x 3x

2 x

3 2

∫

_= − + =



 ₋

− =

    

  

= =

= e ·dx

9 2 9

e · x 2 3

e · x dx · e 3 1 e 3 1 · x · 3 2 3

e · x e 3 1 v ; dx · e dv

dx du ; x u

x 3 x

3 x

3 2 x

3 x

3 2 x 3 x

3

C ) 2 x 6 x 9 ·( 27 e C e 3 1 · 9 2 9

e · x 2 3

e ·

x2 3x ₋ 3x ₊ 3x ₊ ₌ 3x 2₋ ₊ ₊

=

Un tipo especial de integrales por partes son las de función exponencial por función trigonométrica, cuya principal propiedad es que son iterativas.

Ejemplo 4.

(

)

∫

= − − =

    

  

= =

− = =

= e · senx·dx

2 1 e 2 1 · x cos e

2 1 v ; dx · e dv

dx · x sen du ; x cos u dx · e · x

cos 2x 2x 2x 2x 2x

=     

  

= =

= +

=

∫

2x 2x 2x

x 2

e · 2 1 v ; dx · e dv

x·dx cos du x; sen u dx · e · x sen 2 1 2

e · x cos

∫

= + −

  

 ₋

+

= e ·cosx·dx

4 1 4

x n se e 2

x ·cos e dx · x ·cos e · 2 1 e

· 2 1 · x sen · 2 1 2

x ·cos

e2x 2x 2x 2x 2x 2x

igualando con la integral inicial, queda:

∫

= + − · e cosx·dx

4 1 4

x sen · e 2

x ·cos e dx · x cos

e 2x

x 2 x

2 x

2

Pasando la integral del segundo miembro al primero, sumando y despejando queda:

∫

+

    

  

+

= C

4 x sen · e 2

x ·cos e 5 4 dx · x cos

e2x 2x 2x

expresión que simplificada queda de la forma:

(

)

∫

= ·2cosx+senx +C

5 e dx · x cos

(3)

INTEGRALES POR PARTES

Resolverlas siguientes integrales

1.

∫

x⋅sec2x⋅dx

2.

∫

eaxcosbx⋅dx

3.

∫

(

x²+1

)

⋅2xdx

4.

∫

x²⋅e4x⋅dx

5.

∫

(

Lnx

)

2dx

6.

∫

x

(

log3x

)

2dx

7.

∫

⋅

− ⋅

dx ² x 1

x arcsen x

8.

∫

(

x²+1

)

⋅sen2x⋅dx

9.

∫

(

x²⋅Lnx−x⋅Ln2x

)

⋅dx

10. dx

∫

arcsen x⋅

11.

∫

2x⋅actg x⋅dx

12.

∫

(

x₋x2

)

·ex3_⋅dx

13.

∫

⋅dx x

x Ln

14.

∫

x·cos2x⋅dx

(4)

INTEGRALES POR PARTES

Resolverlas siguientes integrales

1.

∫

= ⋅ −

∫

⋅ = ⋅ −

∫

⋅ =

   

 

= ⇒ ⋅ =

= ⇒ = =

⋅

⋅ dx

x cos

x sen x tg x dx x tg x tg x x tg v dx x ² sec dv

dx du x u dx

x ² sec x

( )

−

_∫

− ⋅ = ⋅ + +

− ⋅

= dx x tg x Lncosx C

x cos

x sen 1 x tg x

2. =

( )

⋅ − ⋅

( )

− ⋅ ⋅ =

    

  

= ⇒ =

⋅ ⋅ − = ⇒ =

=

⋅

∫

e b senbx dx

a 1 e a 1 bx cos e

a 1 v e dv

dx bx sen b du bx cos u dx bx cos

eax ax ax ax ax

( )

=

    

  

= ⇒ =

⋅ ⋅

= ⇒ =

= ⋅ ⋅

+ ⋅

= ax

∫

ax ax _eax

a 1 v e dv

dx bx cos b du bx sen u dx bx sen e a b e bx cos a 1

( )

⋅ + ⋅_

( )

−

( )

⋅ _=

=

∫

e b·cosbx dx

a 1 e a 1 bx sen a b e bx cos a

1 ax ax ax

( )

⋅ + ⋅

( )

⋅ −

_∫

( )

⋅

= e cosbx dx

² a

² b e bx sen ² a

b e bx cos a

1 ax ax ax

igualando con la integral inicial

( )

_∫

( )

∫

⋅ = ⋅ + ⋅ ⋅ − e cosbx ⋅dx

² a

² b e bx sen ² a

b e bx cos a 1 dx bx cos

eax ax ax ax

ecuación de la que se puede despejar el valor de la integral

( )

ax

( )

ax

( )

ax

ax _sen_bx _e

² a

b e bx cos a 1 dx bx cos e ² a

² b dx bx cos

e ⋅ +

∫

⋅ = ⋅ + ⋅ ⋅

∫

( )

ax

( )

ax

ax _sen_bx _e

² a

b e bx cos a 1 dx bx cos e ² a

² b

1  ⋅ = ⋅ + ⋅ ⋅

   

 +

_∫

( )

      +

⋅ ⋅

+ ⋅ =

⋅

∫

² a

² b 1

e bx sen ² a

b e bx cos a 1 dx bx cos e

ax ax

ax

operando y ordenando

( )

(

acos

( )

bx b sen

( )

bx

)

C ²

b ² a

e dx bx cos e

ax

ax ₊ _⋅ ₊

+ = ⋅

∫

3.

(

)

=

(

+

)

⋅ − ⋅ ⋅ =

    

  

= = ⇒ =

⋅ = ⇒ + = = ⋅

+

∫

2x dx

2 Ln

2 2 Ln

2 1 ² x 2 Ln

2 v v dx 2 dv

dx x 2 du 1 ² x u dx 2 1 ²

x x x x x x

(

)

₌

    

  

= ⇒ =

= ⇒ = = ⋅ ⋅ ⋅ − ⋅ +

=

∫

2 Ln

2 v dx 2 dv

dx du x u dx 2 x 2 Ln

2 2

Ln 2 1 ²

x x

x

(

)

(

)

∫

_= + ⋅ − ⋅ + ⋅ ⋅ =

  

  

⋅ − ⋅ ⋅ − ⋅ +

= 2 dx

2 Ln

2 2 Ln

2 x 2 2 Ln

2 1 ² x dx 2 Ln

2 2 Ln

2 x 2 Ln

2 2

Ln 2 1 ²

x x

2 2

x x

x

(

)

(

)

⋅ ₊ ₌

+ ⋅ − ⋅ + = + ⋅ + ⋅ − ⋅ +

= C

2 Ln

2 2 2 Ln

2 x 2 2 Ln

2 1 ² x C 2 Ln

2 2 Ln

2 x 2 2 Ln

2 1 ² x

3 x 2

x x

x 2 2

(5)

(

)

_C 2 Ln 2 2 Ln x 2 2 Ln 1 ² x

2x ₂ ₃ +

     + ₋ ₊ ⋅ =

4. = ⋅ − ⋅ ⋅ =

        = ⇒ ⋅ = ⋅ = ⇒ = = ⋅ ⋅

∫

e 2x dx

4 1 e 4 1 ² x e 4 1 v dx e dv dx x 2 du ² x u dx e ²

x 4x 4x 4x 4x 4x

=       _⋅ ₋ _⋅ ⋅ − =         = ⇒ ⋅ = = ⇒ = = ⋅

−

∫

e dx

4 1 e 4 1 x 2 1 e 4 ² x e 4 1 v dx e dv dx du x u dx e x 4 2 e 4 ²

x 4x 4x 4x

x 4 x 4 x 4 x 4 = + ⋅ + ⋅ − = ⋅ + ⋅ −

=

∫

e C

4 1 8 1 e 8 x e 4 ² x dx e 8 1 e 8 x e 4 ²

x 4x 4x 4x 4x 4x 4x

(

8x² 4x 1

)

C 32 e C e 32 1 e 8 x e 4 ²

x 4x ₋ _⋅ 4x₊ _⋅ 4x ₊ ₌ 4x _⋅ ₋ ₊ ₊

=

5.

(

)

= ⋅ − ⋅ ⋅ ⋅ =

        = ⇒ = ⋅ = ⇒ = =

∫

dx x 1 x Ln 2 x x x ² Ln x v dx dv x 1 x Ln 2 du x ² Ln u dx x Ln 2 =       _⋅ ₋ _⋅ _⋅ ⋅ − ⋅ =         = ⇒ = ⋅ = ⇒ = = ⋅ ⋅ − ⋅

=

∫

dx

x 1 x x x Ln 2 x ² Ln x x v dx dv dx x 1 du x Ln u dx x Ln 2 x ² Ln x C x 2 x Ln x 2 x ² Ln x dx 2 x Ln x 2 x ² Ln

x⋅ − ⋅ + = ⋅ − ⋅ + +

=

∫

6.

(

)

(

)

=

          = ⇒ ⋅ = ⋅ ⋅ ⋅ ⋅ = ⇒ = =

∫

2 ² x v dx x dv dx 3 Ln x 1 x log 2 du x log u dx x log x 3 2 3 2 3

(

)

⋅ = ⋅ ⋅ ⋅ ⋅ − ⋅

=

∫

dx

3 Ln x 1 x log 2 2 ² x 2 ² x x

log3 2 3

(

)

₌           = ⇒ ⋅ = ⋅ ⋅ = ⇒ = = ⋅ ⋅ − ⋅ =

∫

2 ² x v dx x dv dx 3 Ln x 1 du x log u dx x log x 3 Ln 1 2 x log ² x 3 3 2 3

(

)

(

)

∫

+ ⋅ = ⋅ ⋅ − ⋅ =     _⋅ ⋅ ⋅ − ⋅ − ⋅

= x dx

3 ² Ln 2 1 3 Ln 2 x log ² x 2 x log ² x dx 3 Ln x 1 2 ² x 2 ² x x log 3 Ln 1 2 x log ²

x ₃ 2 ₃

3 2

3

(

)

_C

3 ² Ln 2 1 3 Ln x log x log 2 ² x C 2 ² x 3 ² Ln 2 1 3 Ln 2 x log ² x 2 x log ²

x 2 3

3 3 2 3 ₊       ⋅ + − ⋅ = + ⋅ + ⋅ ⋅ − ⋅ =

7.

(

)

=

− ⋅ − − − − − ⋅ =             − − = ⇒ ⋅ − = − = ⇒ = = ⋅ − ⋅

∫

arcsenx 1 x² 1 x² ₁dx_x_²

² x 1 v dx ² x 1 x dv ² x 1 dx du x arcsen u dx ² x 1 x arcsen x C x ² x 1 x arcsen dx ² x 1 x

arcsen ⋅ − + =− ⋅ − + +

−

(6)

8.

(

)

(

)

− − ⋅ ⋅ =     − ⋅ + =         − = ⇒ ⋅ = ⋅ = ⇒ + = = ⋅ ⋅ +

∫

cos2x 2x dx

2 1 x 2 cos 2 1 1 ² x x 2 cos 2 1 v dx x 2 sen dv dx x 2 du 1 ² x u dx x 2 sen 1 ² x

(

)

₌         = ⇒ ⋅ = = ⇒ = = ⋅ ⋅ + + −

=

∫

_sen₂_x

2 1 v dx x 2 cos dv dx du x u dx x 2 cos x 2 x 2 cos 1 ² x

(

)

(

)

_C

4 x 2 cos 2 x 2 sen x 2 x 2 cos 1 ² x dx x 2 sen 2 1 x 2 sen 2 1 x 2 x 2 cos 1 ²

x ⋅ ₊ ₊

+ + − = ⋅ − ⋅ + + − =

∫

9.

∫

(

x²⋅Lnx−x⋅Ln2x

)

⋅dx=

∫

x²⋅Lnx⋅dx−

∫

x⋅Ln2x⋅dx

∫

= ⋅ − ⋅ ⋅ = ⋅ − ⋅ =           = ⇒ ⋅ = ⋅ = ⇒ = = ⋅

⋅ x² dx

3 1 3 x Ln ³ x dx x 1 3 ³ x 3 ³ x x Ln 3 ³ x v dx ² x dv dx x 1 du x Ln u dx x Ln ² x 1 C 9 ³ x 3 x Ln ³ x C 3 ³ x 3 1 3 x Ln ³

x ⋅ ₋ ₊

= + ⋅ − ⋅ = = ⋅ ⋅ ⋅ ⋅ − ⋅ =           = ⇒ ⋅ = ⋅ ⋅ ⋅ = ⇒ = = ⋅ ⋅

∫

dx x 1 x Ln 2 2 ² x 2 ² x x Ln 2 ² x v dx x dv dx x 1 x Ln 2 du x Ln u dx x Ln x 2 2 2 =     _⋅ ₋ _⋅ _⋅ − ⋅ =           = ⇒ ⋅ = ⋅ = ⇒ = = ⋅ ⋅ − ⋅

=

∫

dx

x 1 2 ² x 2 ² x x Ln 2 x Ln ² x 2 ² x v dx x dv dx x 1 du x Ln u dx x Ln x 2 x Ln ²

x 2 2

2 2 2 C 4 ² x 2 x Ln ² x 2 x Ln ² x dx x 2 1 2 x Ln ² x 2 x Ln ²

x ⋅ ₊ ₊

− ⋅ = ⋅ + ⋅ − ⋅

∫

sustituyendo en la integral inicial:

(

)

∫

⋅ − ⋅ ⋅ = ⋅ − − ⋅ + ⋅ − +C

4 ² x 2 x Ln ² x 2 x Ln ² x 9 ³ x 3 x Ln ³ x dx x Ln x Lnx ²

x 2 2

10. = − ⋅ − ⋅ =         = → = − = → = = ⋅

∫

2 ₁ _x2

dx x arcsen x x x v dx

dv 1 x

dx du arcsen x u dx sen x arc

( )

−

(

−

)

⋅ = ⋅ −−

( )

− + = − − ⋅ = ⋅ − − − − ⋅

=

∫

− C

2 1 x 1 2 1 arcsen x x dx x 2 x 1 2 1 arcsen x x dx x 1 x 2 2 1 arcsen x x 2 1 2 2 1 2 2 C x 1 arcsen x

x⋅ + − 2 +

=

11. =

+ − = + ⋅ − =         = → ⋅ = + = → = = ⋅ ⋅

∫

dx x 1 x x arctg x x 1 dx x x arctg x x v dx x 2 dv x 1 dx du x arctg u dx x actg x

2 2 2 ₂ 2 2₂

2 2

( )

x 1 x C x arctg C x arctg x x arctg x dx x 1 1 1 x arctg

x2 ₂_ = 2 − + + = ⋅ 2+ − +

(7)

12.

(

)

(

)

=

(

−

)

⋅ − ⋅

(

−

)

⋅ = 

   

  

= → ⋅ =

⋅ − = → − = = ⋅

−

∫

x x 3e 3e 1 2x dx

e 3 v dx e dv

dx x 2 1 du x x u dx e · x

x 2 x3 x3

3 x 3

x 2 3

x 2

(

)

(

)

(

)

(

)

( )

_=

  

 

⋅ − −

⋅ − − − =

   

 

= → ⋅ =

− = → − = = ⋅ ⋅ − − −

=

∫

3e x x 3 1 2x 3e

∫

3e 2 dx

e 3 v dx e dv

dx 2 du x 2 1 u dx e x 2 1 3 x x e

3 x3 2 x3 x3

3 x 3

x 3

x 2

3 x

(

−

)

−

(

−

)

− ⋅ =

(

−

)

−

(

−

)

− ⋅ + =

=3ex3 x x2 9ex3 1 2x 18

∫

ex3 dx 3ex3 x x2 9ex3 1 2x 18 3ex3 C

(

x 7x 21

)

C e

3 x3 2₋ ₊ ₊

− =

13. = ⋅ − ⋅ = ⋅ − =

      

      

= = → ⋅ = =

= → =

=

⋅

∫

− _x

dx 2 x Ln x 2 x dx x 2 x 2 x Ln x 2 2 1 x v dx x x dx dv

x dx du x Ln u dx

x x Ln

2 1 2

1

(

−

)

+ =

⋅ = + − ⋅ = + −

⋅ = −

⋅

=

∫

− C 2 x Ln x 4 x C 2 x Ln x 2 C

2 1 x 2 x Ln x 2 dx x 2 x Ln x

2 2

1 2

1

14. ⋅ =

  

  + −    

  + =     

  

+ = → ⋅ =

= → = =

⋅

∫

dx

4 x 2 sen 2 x 4

x 2 sen 2 x x 4

x 2 sen 2 x v dx x cos dv

dx du x u dx

x ·cos

x 2 2

C 8

x 2 sen 4

x 2 cos x 4 x C 8

x 2 cos 4 x 4

x 2 sen x 2

x2 ₊ ⋅ ₋ 2 ₊ ₊ ₌ 2 ₊ ⋅ ₋ ₊

=

15. =

   

 

= → =

= → = = ⋅ = ⋅ ⋅ =

    

  

⋅ =

= + =

⋅

∫

+

t t

2 1

x

e v dt e dv

dt du t u dt e t 2 dt t 2 e dt t 2 ds

t 1 x dx

e 2

= + −

⋅ + =     

  

+ =

= + = + − ⋅ = −

⋅

=

∫

2 x 1 e + 2e + C

1 x t

t 1 x C e 2 e t 2 dt e 2 e t

2 t t t t 2 x 1 x 1

(

x 1 1

)

C e

2 x 1 + − +

= +

(8)