Page 1

INTEGRALES INDEFINIDAS RESUELTAS

RELACIÓN DE EJERCICIOS DE INTEGRALES.

1.

∫x

5

∫ 2.

dx

x 5 dx =

∫( x +

x 5+1 x6 +k = +k 5 +1 6

x )dx

( x + x )dx =

( x + x 1 / 2 )dx =

=

3.

 3 x x   x − 4 

5.

Sol: 6 x −

∫ (3 x

1 2

x dx = x

∫x

4  1  + 2  dx  2 + x x x 

4  1  + 2  dx =  2+ x x x 

x

=

6.

1 2

dx x

4

1 2

dx =

−1

3 2

x dx =

−2

1 2

x5 1 2 +k =6 x − x x +k 5 10

3 +1 2

2 2 x x +k 5

5 2

x x 2 x5 2x2 x +k = +k = +k = +k 3 5 5 5 +1 2 2

Sol: −

∫(x

3 +1 2

5 2

1 x 1 x x 1 x x )dx = 3 ⋅ − ⋅ + k = 3⋅ − ⋅ +k = 1 3 1 4 4 4 5 − +1 +1 2 2 2 2

Sol: 2

1 2 x x +k 10

1 − +1 2

3 2

x 2 dx x 2

3 2

x x x x x2 2 x3 + +k = + +k = + +k = 3 1+1 1 2 2 3 +1 2 2

  dx  

 3 x x   dx = −  x  4  

2

x2 2x x + +k 2 3

=6 x −

4.

1 +1 2

1+ 1

+ 4x

3 2

1 8 − + 2x + k x x 3 − +1

x − 2 +1 x 2 + 2)dx = +4 + 2x + k = 3 −2+1 − +1 2

1 2

x x 1 1 1 8 +4 + 2x + k = − − 8 ⋅ 1 + 2x + k = − − + 2x + k 1 −1 x x x − x2 2

Sol:

1

44 3 x +k 3


INTEGRALES INDEFINIDAS RESUELTAS

7.

1 − +1

1 4

dx = 4 x

1   2  x + 3  dx x 

3

3

x 4 x4 4x 4 4 x dx = +k = +k = +k = 1 3 3 3 − +1 4 4 −

2

∫  x 

2

+

3

2

1   x

4

x3 + k

x5 3 2 3 2 + x x + 33 x + k 5 4

Sol:

8

2

8

1

1 3 1 3 = ⋅ x 5 + ⋅ x 3 + 3. x 3 + K = ⋅ x 5 + ⋅ 3 x 8 + 3.3 x + K = 5 4 5 4

=

8.

∫ ∫

9.

∫ 10.

Ln( x ) ⋅ dx = x 2

Ln( x ) ⋅

1  ⋅ dx =  x 

f

α

Sol:

x ⋅dx

1 2 L x +k 2

2  (Ln( x )) ⋅ f ' dx  = +k 2 

1 ⋅ tg 2 x + k 2

2  (tg x ) f 1 ⋅ f ' dx  = +k 2 

 tg x ⋅ sec 2 x ⋅ dx =  

Sol:

 sen 2 x ⋅ cos x ⋅ dx ==  

sen 3 x +k 3

3  sen x f 2 ⋅ f ' dx  = +k 3 

Sol: −

cos 3 x ⋅sen x ⋅dx

∫ 12.

Sol:

sen 2 x ⋅cos x ⋅dx

∫ 11.

1 5 3 23 2 ⋅ x + ⋅ x . x + 3. 3 x + K 5 4

Lx ⋅ dx x

∫tg x ⋅sec

cos 4 x +k 4

cos 4 x cos x ⋅ sen x ⋅ dx = − cos x ) ⋅ dx = +k    x ⋅ (− sen   4 3 f f'

∫x

1

1 5 2  −  − x5 x3 x3 dx =  x 2 + x 3  dx = ( x 4 + 2 x 3 + x 3 ).dx = + 2⋅ + +K=   8 1 5   3 3

3

x 2 +1 ⋅dx

3

Sol:

2

1 ( x 2 + 1) 3 + k 3


INTEGRALES INDEFINIDAS RESUELTAS

∫ 13.

∫ ∫ 14.

15.

16.

2

2

xdx 2x + 3

xdx 2x + 3 2

=

xdx 2x 2 + 3

=

1 2

 f'  = dx = 2 2 f  2 2x + 3 

1 1 ( 2 x + 3) 1 4x ⋅ ( 2 x 2 + 3) dx = ⋅ +k= 2x2 + 3 + k      1 4 f' 4 2 f −1 / 2 2 2 4

Sol:

x +1

x 2 dx

x +1 3

x +1 3

=

∫ x (x

=

2 3

2

1 2

+ 1) dx =

3

3

2 x3 + 1 + k 3 1 2

  2 f + k =   3

3

x3 +1 + k

cos x ⋅dx sen 2 x

cos x sen − 1 x 1 −2 ⋅ dx = cos x ⋅ sen x dx = + k = − +k     2 −1 sen x sen x f' f−2

Sol: −

1 +k sen x

∫x ⋅( x

2

Sol:

+1) 4 dx

( x 2 + 1) 5 +k 10

1 1 ( x 2 + 1) 5 ( x 2 + 1) 5 2 4 x ⋅ ( x + 1) dx = 2x ⋅ ( x + 1) dx = ⋅ +k= +k 2 f '    4  2 5 10 2

4

∫ cos

sen x ⋅ dx 3 x

1 2

1 1 ( x + 1) 2 3 x 2 ( x 3 + 1) dx = ⋅ +k = x3 + 1 + k 1 3 3 3 2

 f'  = dx = 2 x +1   2 f 3 x 2 dx

o

  1 f + k = 2x2 + 3 + k 2  

4 xdx

3

x dx

1 2

2

x 2 dx

2

1 2x2 + 3 + k 2

Sol:

2

∫ 17.

3 2

( x 2 + 1) 3 1 1 ( x + 1) 2 x x + 1 ⋅ dx = 2x ⋅ ( x + 1) dx = ⋅ +k= +k 3 2 f '   1 / 2 2 3 f 2 1 2

f

Sol:

1 +k 2 cos 2 x

sen x cos − 2 x 1 −3 ⋅ dx = − − sen x ⋅ cos x dx = − + k = +k       3 2 −2 cos x 2 cos x f' f −3

3


INTEGRALES INDEFINIDAS RESUELTAS

18.

19.

20.

tg x ⋅ dx cos 2 x

tg x 1 tg 2 x ⋅ dx = tg x ⋅ cos 2 x ⋅ dx = 2 + k cos 2 x    f1

cotg x ⋅ dx sen 2 x

cotg x −1 cotg 2 x ⋅ dx = − cotg x ⋅ ⋅ dx = − +k 2 sen 2 x sen 2 x

1 ⋅ dx cos x tg x − 1

22.

23.

tg 2 x +k 2

f'

Sol: −

cotg 2 x +k 2

Sol:

2

∫ cos 21.

Sol:

2

1 ⋅ dx = x tg x − 1

∫ cos

1 2

x

⋅ (tg x − 1)

1 2

2 tg x −1 +k

⋅ dx =

( tg x − 1) + k = 2 tg x − 1 + k 1 2

L ( x + 1) dx x +1

L ( x + 1) 1 L2 ( x + 1) dx = L ( x + 1) ⋅ dx = +k     x + 1 x+1 2  1

Sol:

f

cos x dx 2sen x + 1

cos x 1 dx = 2 2sen x + 1

sen 2 x dx (1 + cos 2 x ) 2

sen 2 x 1 dx = − 2 2 (1 + cos 2 x ) =

L2 ( x + 1) +k 2

f'

Sol:

1 2

2sen x +1 + k

1

1 2

1 ( 2sen x + 1) 2 2 x + 1) dx = ⋅ + k = 2sen x + 1 + k  cos   x (2sen      1 2 f' −1 / 2 f 2 −

Sol:

1 +k 2(1 + cos 2 x )

1 (1 + cos 2 x ) −1 − 2sen 2 x ⋅ (1 + cos 2 x ) −2 dx = − ⋅ +k = 2 −1

1 +k 2(1 + cos 2 x )

4


INTEGRALES INDEFINIDAS RESUELTAS

24.

sen 2 x 1 + sen 2 x

sen2 x 1 + sen 2 x

dx

Sol: 2 1 + sen 2 x + k

dx =

1

sen2 x (1 + sen 2 x ) 2 dx =

1

2sen sen 2 x ) 2 dx =   x ⋅ cos   x (1 +      f'

f

−1 / 2

1

(1 + sen 2 x ) 2 = + k = 2 1 + sen 2 x + k 1 2

25.

tg x + 1

2

cos x

tg x + 1

∫ 26.

2

cos x

⋅ dx

Sol:

⋅ dx =

∫ (2 + 3sen 2 x )

cos 2 x

3

1 2

(tg x + 1) ⋅

∫ ∫

28.

29.

sen 3 x 4

cos 3 x

Ln 2 x dx x

Ln 2 x dx = x

arc sen x dx 1 − x2

3

1 +k cos 3 x −

4 3

1

1 cos 3 3 x − 3sen 3 x cos 3 x dx = − ⋅ +k =          1 3 f' f −4 / 3 − 3 −

Sol:

1 ( 2 + 3sen 2 x ) −2 ⋅ +k = 6 −2

1 1 +k 12 ( 2 + 3sen 2 x ) 2

Sol:

1 dx = − 4 3 cos 3 x

1 1 +k 12 ( 2 + 3sen 2 x ) 2

6 cos 2 x ⋅ ( 2 + 3sen 2 x ) −3 dx =

dx

sen 3 x 3

1 (tg x + 1) 2 ⋅ dx = +k = (tg x + 1) 3 + k 2 3 3 cos x 2

Sol: −

cos 2 x 1 dx = 3 6 ( 2 + 3sen 2 x )

3

3 2

dx

=−

27.

2 (tg x + 1) 3 + k 3

Ln 3 x +k 3

1 Ln 3 x Ln x dx = +k x 3 2

Sol: 5

arc sen 2 x +k 2

3

1 +k cos 3 x


INTEGRALES INDEFINIDAS RESUELTAS

∫ 30.

31.

32.

33.

1 − x2

arc cos 2 x dx

arc tg x dx 1 + x2

= − arc cos 2 x

1− x2

arc tg x dx = 1 + x2

−1 1 − x2

arc sen 2 x +k 2

Sol: −

arc cos 3 x +k 3

dx = −

arc cos 3 x +k 3

Sol:

arc tg x ⋅

arc tg 2 x +k 2

1 arc tg 2 x dx = +k 2 1 + x2

arc ctg x dx 1 + x2

arc ctg x −1 arc ctg 2 x dx = − arc ctg x ⋅ dx = − +k 2 1 + x2 1 + x2

x dx x +1

Sol: −

arc ctg 2 x +k 2

Sol:

2

∫ ∫

35.

1− x2

dx =

arc cos 2 x dx

∫x 34.

1

= arc sen x

1− x2

arc sen x dx

2

x 1 dx = 2 +1

∫x

 2x dx =  2 +1 

f'

Sol:

 dx −1 =− ⋅ dx =  1−x 1−x 

∫ f dx =Ln f + k  = 2 Ln( x

dx 1−x

1 Ln( x 2 + 1) + k 2

1

2

+ 1) + k

−Ln 1 −x +k

 f' dx =Ln f + k  = −Ln 1 − x + k f 

dx 3 x −7

Sol: 

1 L 3x − 7 + k 3 

∫ 3 x − 7 = 3 ∫ 3 x − 7 ⋅ dx = ∫ f dx =Ln f + k  = 3 Ln 3 x − 7 + k dx

36.

1

3

f'

1

1 2

dx 5 −2x

Sol: − L 5 − 2 x + k 

∫ 5 − 2 x = − 2 ∫ 5 − 2 x ⋅ dx = ∫ f dx =Ln f + k  = − 2 Ln 5 − 2 x + k dx

1

−2

f'

6

1


INTEGRALES INDEFINIDAS RESUELTAS

37.

38.

39.

∫x

2

x +1 dx + 2x + 3

x+1 1 dx = 2 x + 2x + 3

dx x ⋅ Ln x

dx = x ⋅ Ln x

2

2( x + 1) 1 dx = 2 x + 2x + 3 2

1 dx  x = Ln x 

Ln Ln x +k

 f' dx = Ln | f | + k  = Ln | Ln x | + k f 

∫tg x dx

Sol: − sen x

sen x

−Ln cos x +k

⋅ dx = −Ln cos x + K

1 2

∫tg 2 x dx

∫ 41.

2x + 2 1 dx = L x 2 + 2 x + 3 + k 2 x + 2x + 3 2

Sol:

∫tg x ⋅ dx = ∫ cos x ⋅ dx = − ∫ cos x 40.

1 L x2 + 2x + 3 + k 2

Sol:

Sol: − L cos 2 x + k

tg 2 x ⋅ dx =

sen 2 x 1 ⋅ dx = − cos 2 x 2

− 2sen 2 x 1 ⋅ dx = − Ln cos 2 x + K cos 2 x 2

∫ctg x dx

Sol:

Ln sen x +k

Sol:

1 L sen (5 x − 7 ) + k 5

∫ctg x ⋅ dx = ∫ sen x ⋅ dx = Ln sen x + K cos x

42.

∫ctg (5 x −7) dx

∫ 43.

Sol: − L cos 3 x + k

dx = ctg 3 x

tg 3 x ⋅ dx =

x

∫(ctg e

x

sen 3 x 1 ⋅ dx = − cos 3 x 3

∫ctg 3 dx x

5 cos(5 x − 7 ) 1 ⋅ dx = Ln sen (5 x − 7 ) + K sen (5 x − 7 ) 5 1 3

dx

∫ ctg 3 ⋅ dx = ∫ 45.

cos(5 x − 7 ) 1 ⋅ dx = sen (5 x − 7 ) 5

∫ctg 3 x ∫

44.

ctg (5 x − 7 ) ⋅ dx =

− 3sen 3 x 1 ⋅ dx = − Ln cos 3 x + K cos 3 x 3 x

Sol: 3L sen 3 + k x 3 ⋅ dx =3 x sen 3 cos

)e x dx

1 x cos 3 3 ⋅ dx = 3 ⋅ Ln sen x + K x 3 sen 3

Sol:

7

L sen e x +k


INTEGRALES INDEFINIDAS RESUELTAS

∫ 46.

(ctg e x )e x ⋅ dx =

(cos e x )e x ⋅ dx = Ln sen e x + K sen e x

x   tg 4 x − ctg  dx 4 

x   tg 4 x − ctg  dx = 4 

1 4

Sol: Sol : − Ln cos 4 x − 4Ln sen

=−

47.

1 4

x   sen 4 x cos  4  dx =  − x  cos 4 x sen    4

− 4sen 4 x dx − 4 cos 4 x

∫ 2sen x + 3 dx cos x

Sol:

∫ ∫ dx 48. ∫ (1 + x ) arc tg x cos x 1 dx = 2sen x + 3 2

Sol:

49.

50.

dx = (1 + x ) arc tg x 2

∫ cos

2

∫ ∫

1 − x arc sen x

1 x cos 4 4 dx = − 1 Ln cos 4 x − 4 Ln sen x + k x 4 4 sen 4

1 Ln ( 2sen x + 3) + k 2

Ln arc tg x +k

1 1 + x 2 dx = Ln arc tg x + k arc tg x

dx x ( 3tg x + 1)

dx = 2 cos x ( 3tg x + 1)

x 4 dx = x sen 4 cos

2 cos x 1 dx = Ln ( 2sen x + 3) + k 2sen x + 3 2

2

sen 4 x dx − cos 4 x

Sol:

1 cos 2 x dx = 1 3tg x + 1 3

dx

1 Ln ( 3tg x + 1) + k 3

3 cos 2 x dx = 1 Ln ( 3tg x + 1) + k 3tg x + 1 3

Sol:

2

Ln arc sen x +k

1

∫ 51.

dx 1 − x 2 arc sen x

=

1− x2 dx = Ln arc sen x + k arc sen x

cos 2 x dx 2 + 3sen 2 x

Sol:

1 Ln 2 + 3sen 2 x + k 6

∫ 2 + 3sen 2 x dx = 6 ∫ 2 + 3sen 2 x dx = 6 Ln 2 + 3sen 2 x + k cos 2 x

1

6 cos 2 x

1

8

x +k 4


INTEGRALES INDEFINIDAS RESUELTAS

52.

∫e

∫e 53.

∫e

sen x

x

f

 1 dx = e f + k  = e 2 x + k  2

x2

x

 f ' e dx = e + k  = 2e 2 + k  f

f

Sol: e sen x + k

∫f ' e

 cos x dx = 

sen x

∫a

dx =e

f

f

 +k =e sen x +k  2

ax +k Sol: 2L a

⋅ x ⋅ dx 1 2Ln a

2

a x x ⋅ dx =

2 1 x2 a x 2 x ⋅Ln  a dx = 2Ln a a + k x2

D( a

)

x

x

Sol: ae a + k

e a dx x

e a dx = a

∫(e

∫e

2x

)

2

− 3x

x

x

e

Sol:

dx

dx =

∫e

5x

4x

dx =

1 4

∫ 4e

4x

dx =

x

1 4x e +k 4

1 4x e +k 4

1 3

Sol: − e −3 x + k

dx

1 3

1 ( −3)e −3 x dx = − e −3 x + k 3

Sol:

dx

5 x e x dx =

∫(e

x

1 a e dx = ae a + k a

e −3 x dx = −

∫5

∫ 60.

∫ f 'e

 dx =  

cos x dx

2x 2

59.

2x

x

∫ (e ) 58.

∫ 2e

1 2x e +k 2

Sol: 2e 2 + k

x 2

∫ 57.

1 2

1 2  e dx = 2 e dx =  2 

∫ 56.

dx =

x

∫e 55.

2x

Sol:

dx

e 2 dx

∫ 54.

2x

∫ )

(5e ) x dx =

+a 5 x dx

1 Ln(5e )

5x e x +k Ln 5 + 1

(5e ) x Ln( 5e )dx =

Sol: 9

1 5x e x ( 5e ) x + k = +k Ln(5e ) Ln 5 + 1

1  5x a5x  e + +k 5  L a 


INTEGRALES INDEFINIDAS RESUELTAS

∫ (e

5x

)

+ a 5 x dx =

1 5

=

61.

62.

∫e

x2 + 4 x+ 3

(a x − b x ) 2 dx a xbx

(a x − b x ) 2 dx = a xb x

=

+4 x +3

1 5 Ln a

5a 5 x Ln a d x =

1 5x 1 e + ⋅ a5x + k = 5 5 Ln a

1  5x a5x  e + +k 5  L a 

Sol:

( x +2) dx

ex

2

5e 5 x dx +

( x + 2) dx =

1 2

ex

2

+4 x +3

1 x 2 +4 x + 3 e +k 2

2( x + 2) dx =

1 x 2 +4 x +3 e +k 2 x

x

a b   −  Sol:  b   a  − 2x + k La −Lb

a 2 x − 2a x b x + b 2 x dx = a xbx

 a2x  b2x  x x + x x − 2  dx = a b a b 

 ax bx   x + x − 2  dx = a b 

x x  a  x  b  x  1 1 a b ⋅  + ⋅   − 2x + k =   +   − 2 dx = a b b a a  b   Ln    Ln    b a x

x

1 1 a  b = ⋅  + ⋅  − 2x + k = Ln a − Ln b  b  Ln b − Ln a  a  x

x

a b     b  a = − − 2x + k = Ln a − Ln b Ln a − Ln b

63.

64.

ex dx 3 + 4e x

ex 1 dx = x 4 3 + 4e

x

Sol:

x

a b   −  b  a  − 2x + k La −Lb

1 Ln ( 3 + 4e x ) + k 4

4e x 1 dx = Ln ( 3 + 4e x ) + k x 4 3 + 4e

∫cos 5 xdx

Sol:

1 sen 5 x + k 5

∫cos 5 xdx = 5 ∫5 cos 5 xdx = ∫ f ' ( x ) ⋅ cos f ( x )dx = sen f ( x ) + k  = 5 sen 5 x + k 1

65.

sen

x dx 3

Sol: − 3 cos

10

x +k 3

1


INTEGRALES INDEFINIDAS RESUELTAS

x dx = 3 3

sen

1 x  sen dx =  3 3 

x +k 3

= −3 cos

66.

∫sec

2

x  f ' ( x ) ⋅ sen f ( x )dx = −cos f ( x ) + k  = 3( −cos ) + k = 3 

Sol:

( 7 x +2)dx

1 tg (7 x + 2) + k 7

∫sec (7 x + 2)dx = 7 ∫7 sec (7 x + 2)dx = ∫ f ' ( x ) sec 1

2

=

67.

∫x cos 3 x

2

2

2

 f ( x )dx = tg f ( x ) + k  = 

1 tg (7 x + 2) + k 7

Sol:

dx

1 sen 3 x 2 + k 6

∫ x cos(3 x )dx = 6 ∫ 6 x cos(3 x )dx = 6 sen (3 x ) + k 1

2

68.

∫tg

2

1

2

2

Sol: tg x − x + k

x dx

Por trigonometría sabemos que tg 2 x + 1 = sec 2 x ⇒ tg 2 x = sec 2 x − 1, entonces

∫tg 69.

70.

cos ( Ln( x) ) dx x

cos ( Ln( x) ) dx = x

∫tg

3

2

∫(sec

x dx =

2

∫sec

x −1)dx =

2

∫dx =tg x −x +k

x dx −

Sol: sen ( Ln(x) ) + k

cos ( Ln( x) )

1 dx = sen ( Ln(x) ) + k x Sol:

x dx

tg 2 x + Ln cos x + k 2

∫ tg x dx = ∫ tg x ⋅ tg x dx = ∫ tg x ⋅ (sec x − 1) dx = ∫ tgx ⋅ sec   x dx − ∫ tg x dx = 3

2

2

2

f1

71.

=

tg 2 x − 2

cos

dx x

x

sen x tg 2 x dx = + cos x 2

f

− sen x tg 2 x dx = + Ln cos x + k cos x 2

Sol:

2 sen

11

x +k


INTEGRALES INDEFINIDAS RESUELTAS

∫ 72.

cos

x

dx = x

x 1−x

1 1 dx = 2 cos x dx = 2 sen x + k x 2 x

cos x

dx

4

1 arc sen x 2 + k 2

Sol:

x  f '(x)  arcsen f (x)+ k  dx = dx =  dx =   = 4 22 2 − arc os f ( x) + k 1− x 1− (x )  1− ( f (x)  

∫ ∫ ∫ x

=

73.

∫ ∫

74.

∫ ∫

1 − 4x

2

2

1 −4x 2

=

dx

=

1 − (2 x ) 2

1 2

9 − 4x2

=

4x2 9(1 − ) 9

=

2 dx 3

a −b x 2

a −b x 2

2

=

dx 2

2

b x a (1 − 2 ) a 2

=

  1  = arc sen (2 x ) + k 2 1 − ( f ( x ))   2 f ' ( x )dx

dx

1 3 = ⋅ 2 3 2  2x  1−   3 

2 dx 3  2x  1−   3 

=

2

 2x  1  = arc sen ( ) + k 2 2 3 1 − ( f ( x ))   f ' ( x )dx

1 bx arc sen ( ) + k b a

Sol:

2

1 = 2 3  2x  3 1−   3 

dx 2

dx

  = 2   2x   1 −   3 

  1 2  = arc sen ( x ) + k 2  

1 2x arc sen ( )+k 2 3

Sol:

dx

2

1 arc sen (2 x ) + k 2

1 − (2 x ) 2

2

dx

1 − ( f ( x ))

  =  

2dx

dx 9 − 4x

f ' ( x )dx

Sol:

2

dx

dx 2

  dx =  1 −( x )   2x

dx

1 = 2

75.

1 2

∫ a 1 −  bx  dx

2

 a 

12

=

1 a

dx  bx  1−   a 

2

=

1 a ⋅ a b

b dx a  bx  1−   a 

2

=


INTEGRALES INDEFINIDAS RESUELTAS

=

76.

77.

78.

79.

80.

81.

1 b

b dx a

ex dx 3 + 4e x

ex 1 dx = x 4 3 + 4e

e 2x dx 2 + e 2x

e 2x 1 dx = 2x 2 2+e

ex dx 1 + e 2x

ex dx = 1 + e 2x

dx 1 + 2x2

dx = 1 + 2x 2

dx 4 + x2

dx = 4 + x2

xdx 4 x + a4

xdx = 4 x + a4

 bx  1−   a 

  =  

2

1 − ( f ( x )) 2

1 Ln ( 3 + 4e x ) + k 4

Sol:

 4e x dx =  x 3 + 4e 

∫ f ( x) dx = 4 Ln (3 + 4e ) + k f '( x)

 2e 2 x dx =  2x 2+e 

x

f '( x)  1 dx  = Ln ( 2 + e 2 x ) + k f ( x)  2

Sol: arc tg (e x ) + k

  ex f '( x) dx =  dx = arc tg f ( x ) + k  = arc tg (e x ) + k x 2 2 1 + (e )  1 + ( f ( x )) 

1 arc tg ( 2 x ) + k 2

Sol:

dx 1 = 2 1 + ( 2 x) 2

  2 dx f '( x) 1 = dx  = arc tg ( 2 x ) + k 2 2 1 + ( 2 x) 2  1 + ( f ( x )) 

1 x arc tg ( ) + k 2 2

Sol:

1

1 Ln ( 2 + e 2 x ) + k 2

Sol:

 bx  1  = arc sen ( ) + k b a  

f ' ( x )dx

dx 1 = 2 4 x 4(1 + ) 4

1 dx 1 1 x 2 = ⋅ 2 = arc tg ( ) + k 2 2 4 2 2 x x 1+  1+  2 2

dx

1 x2 Sol: arc tg ( 2 ) + k 2a 2 a

xdx 1 = 4 4 x a a 4 ( 4 + 1) a

xdx 1 = 4 4 x a 1+ 4 a

1 a2 = ⋅ 2 a4 2  x2  1 +  2  a 

13

xdx

2x dx a2 = 2  x2  1 +  2  a 


INTEGRALES INDEFINIDAS RESUELTAS

1 = 2 2a

82.

cos xdx a + sen 2 x

cos xdx = a + sen 2 x

2x dx 1 x2 a2 = arc tg ( )+k 2 2a 2 a2  x2  1 +  2  a 

2

cos xdx 1 = 2 2 sen x a a 2 (1 + ) 2 a

1 = 2 ⋅a a

83.

∫x

∫x 84.

cos xdx 1 = 2 2 sen x a 1+ a2

1 cos xdx 1 a = 2 a  sen x  1+   a 

dx

dx 1 − Ln 2 ( x )

arccos x − x

arccos x − x 1− x

2

2

dx =

∫ 1 +  sen x  cos xdx 

a

2

=

1 cos xdx 1 sen x a = arc tg ( )+k 2 a a  sen x  1+   a 

Sol: arc sen (Ln( x )) + k

1 − Ln 2 ( x )

1−x

1 sen x arc tg ( )+k a a

Sol:

2

=

1 dx   x = 2 1 − ( Ln ( x ))  

   = arc sen (Ln( x )) + k 2 1 − ( f ( x ))   f ' ( x )dx

1 2

dx

Sol: − (arccos( x )) 2 + 1 − x 2 + k

arccos x 1− x

2

dx +

−x 1− x

2

dx = − arccos     x f1

−1 x 1− 

2

dx +

∫2 1− x − 2x

2

dx =

f'

  =  

85.

x − arctg x dx 1 + x2

x − arctg x dx = 1+ x2

1+ x

x

dx

 f' 1  dx  = − (arccos( x )) 2 + 1 − x 2 + k 2 2 f  

Sol:

∫1+ x

 = 

86.

f 1 ⋅ f ' dx +

x

2

dx −

∫ 1+ x

arctg x

∫ f ( x) dx − ∫ f f '( x)

1

2

dx =

1 2

1 1 Ln(1 + x 2 ) − (arctg x ) 2 + k 2 2

∫1+ x 2x

2

dx −

∫ arctg x 1 + x 1

2

dx =

 1 1 ⋅ f ' dx  = Ln(1 + x 2 ) − (arctg x ) 2 + k 2  2

Sol: 14

4 (1 + x ) 3 + k 3


INTEGRALES INDEFINIDAS RESUELTAS

1+ x dx = x

1 x

1+

x dx = 2

(1 + x ) = 2⋅

3 2

3 2

∫2

+k=

1

(1 + x

x

)

1 2

  dx =   

f

  ⋅ f ' dx  =  

1 2

4 (1 + x ) 3 + k 3

Veamos como podemos realizar esta misma integral por el método de sustitución o cambio de variable. Haciendo el cambio

1+

x =t

, calculamos dx:

1 2 x

dx = dt ⇒ dx = 2 x dt y sustituimos

en nuestra integral:

1+ x dx = x

t ⋅ 2 x dt = 2 x

3

1 2

3

t2 4 4 3 t dt = 2 t dt = 2 + k = t 2 + k = t +k = 3 3 3 2

una vez realizada la integral hay que deshacer el cambio de variable y volver a la variable x, con lo que nos quedará: =

87.

∫ ∫

1 x 1+

x

1 x 1+

x

dx

4 (1 + x ) 3 + k 3

Sol:

dx =

1 x

1 1+

(1 + x ) = 2⋅

x 1 2

1 2

4 1+

dx = 2

∫2

x +k

1 x

(1 +

x

)

1 2

  dx =   

f

1 2

  ⋅ f ' dx  =  

+ k = 4 1+ x + k

Veamos como podemos realizar esta misma integral por el método de sustitución o cambio de variable. Haciendo el cambio

1+

x =t

, calculamos dx:

en nuestra integral:

15

1 2 x

dx = dt ⇒ dx = 2 x dt y sustituimos


INTEGRALES INDEFINIDAS RESUELTAS

1 x 1+ x

dx =

1 ⋅ 2 x dt = 2 x t

1

1

1

− 1 t2 dt = 2 t 2 dt = 2 + k = 4t 2 + k = 4 t + k = una 1 t 2

vez realizada la integral hay que deshacer el cambio de variable y volver a la variable x, con lo que nos quedará: =4 1 +

88.

x

x +k 5

Sol:

x − 1 dx

Hacemos la sustitución x − 1 = t 2

⇒ x = t2 +1

Calculamos la diferencial de x: dx = 2t .dt calcular. Tendremos:

x . x − 1 ⋅ dx =

y sustituimos en la integral que deseamos

t5 t3  ( t 2 + 1). t 2 ⋅ 2tdt = 2 ( t 2 + 1).t 2 .dt =2 ( t 4 + t 2 ).dt = 2 +  + k = 3 5

5

=

89.

∫x(5 x

2

3

2 2 ( x − 1) 2 + ( x − 1) 2 + k 5 3

3

2 5 2 3 2 2 ⋅ t + ⋅ t + k = ⋅ ( x − 1) 2 + ⋅ ( x − 1) 2 + k 5 3 5 3

Sol:

−3) 7 dx

1 ( 5 x 2 − 3) 8 + k 80

Directamente:

x (5 x 2 − 3) 7 dx =

1 10

=

 10 x (5 x 2 − 3) 7 dx =  

2 8  1 ( 5 x − 3) f 7 ⋅ f ' dx  = ⋅ +k = 8  10

1 ( 5 x 2 − 3) 8 + k 80

Por sustitución:

90.

2 Hacemos 5 x − 3 = t ⇒ 10 xdx = dt ⇒ dx =

dt y sustituimos en nuestra integral 10 x

1 t8 1 ⋅ +k = ( 5 x 2 − 3) 8 + k 10 8 80

x(5 x 2 − 3) 7 dx =

xt 7

dt 1 = 10 x 10

x ( 2 x +5) 10 dx

t 7 dt =

1  ( 2 x + 5) 12 5( 2 x + 5) 11  − Sol:  +k 4 12 11 

Por sustitución: Hacemos 2 x + 5 = t ⇒ x =

t −5 1 ⇒ dx = dt y sustituimos en nuestra integral 2 2 16


INTEGRALES INDEFINIDAS RESUELTAS

x( 2 x + 5) 10 dx =

=

91.

∫xe

x

t − 5 10 1 1 ⋅ t ⋅ dt = 2 2 4

( t − 5)t 10 dt =

1 4

∫(t

11

− 5t 10 )dt =

1  t 12 t 11  1  ( 2 x + 5) 12 5( 2 x + 5)11  − 5 ⋅ + k = −    +k 4  12 11  4 12 11 

Sol: e x ( x − 1) + k

dx

Por el método de integración por partes:

∫ 92.

 u = x ⇒ du = dx  x x x x x xe x dx =  x x  = xe − e dx = xe − e + k = ( x − 1)e + k  dv = e dx ⇒ v = e  ∫( x

I =

2

Sol: e x ( x 2 − 5 x + 10) + k

−3 x +5 ) e x dx

Por el método de integración por partes:

 u = x 2 − 3 x + 5 ⇒ du = (2 x − 3)dx I = ( x − 3 x + 5) e dx =  = x x dv = e dx ⇒ v = e   2

x

= ( x 2 − 3 x + 5) e x − e x (2 x − 3) dx = La integral que nos ha quedado es del mismo tipo que la que pretendemos calcular, por lo que nuevamente aplicaremos el método de integración de partes: Hacemos

 u = 2 x − 3 ⇒ du = 2dx   x x   dv = e dx ⇒ v = e 

∫e

I =( x 2 −3 x +5) e x −

x

y sustituimos:

∫2e

 ( 2 x −3)dx =( x 2 −3 x +5) e x −( 2 x −3 e x − 

=( x 2 −3 x +5) e x −( 2 x −3) e x +2

∫e

x

x

dx =( x 2 −3 x +5) e x −( 2 x −3) e x +2e x +k =

[

]

= ( x 2 − 3 x + 5) − ( 2 x − 3) + 2 e x + k = e x ( x 2 − 5 x + 10) + k

93.

 dx  = 

∫x Ln( x ) dx

Sol:

Por el método de integración por partes:

17

1 2 1 x  Ln( x ) −  + k 2  2


INTEGRALES INDEFINIDAS RESUELTAS

1   u = Ln( x ) ⇒ du = dx   1 1 1 x x Ln( x ) dx =  = x 2 Ln( x ) − x 2 ⋅ dx =  2 x  dv = xdx ⇒ v = 1 x 2  2 2  

=

94.

1 2 1 x Ln( x ) − 2 2

1 2 1 1 1  1 x Ln( x ) − ⋅ x 2 + k = x 2  Ln( x ) −  + k 2 2 2 2  2

x dx =

Sol: x ( Ln( x ) − 1) + k

∫Ln( x ) dx Por el método de integración por partes:

1   1  u = Ln( x ) ⇒ du = dx Ln( x) dx =  = x Ln( x ) − x ⋅ dx = x  x  dv = dx ⇒ v = x 

∫dx =x Ln( x ) −x +k =x (Ln( x ) −1) +k

=x Ln( x ) −

95.

∫x sen x dx

Sol: sen x − x cos x + k

Por el método de integración por partes:

 u = x ⇒ du = dx  x sen x dx =   = − x cos x − − cos x dx =  dv = sen x dx ⇒ v = − cos x  ∫cos x dx =−x cos x +sen x +k

=− x cos x +

96.

Sol:

x cos 2 x dx

x2 1 1 + x sen 2 x + cos 2 x + k 4 4 8

Por el método de integración por partes, hacemos u = x ⇒ du = dx y dv = cos 2 xdx Para calcular el valor de v recurrimos a las razones trigonométricas del ángulo mitad y tendremos que cos 2 x = v=

cos 2 xdx =

1 + cos 2 x . Por tanto, 2

1 + cos 2 x 1 dx = 2 2

(1 + cos 2 x )dx =

1 1 ( x + sen 2 x ) 2 2

En consecuencia:

x cos 2 x dx = x ⋅

1 1 ( x + sen 2 x ) − 2 2

1 1 ( x + sen 2 x ) dx = 2 2 18


INTEGRALES INDEFINIDAS RESUELTAS

=

1 2 1 1 ( x + xsen 2 x ) − 2 2 2

∫e

 1 1 1 1  x2 1 sen 2 x )dx = ( x 2 + xsen 2 x ) − ⋅  − cos 2 x  + k = 2 2 2 2  2 4 

1 2 1 x2 1 x2 1 1 x + xsen 2 x − + cos 2 x + k = + x sen 2 x + cos 2 x + k 2 4 4 8 4 4 8

=

97.

(x +

− x

1 −x e (sen x − cos x) + k 2

Sol:

cos xdx

 u = e − x ⇒ du = − e − x dx  − x −x I = e cos xdx =   = e sen x − − e sen xdx =  dv = cos xdx ⇒ v = sen x  −x

∫e

=e −x sen x +

− x

sen xdx

Al aplicar el método de partes nos ha quedado una integral del mismo tipo que la que pretendemos calcular, por lo que volvemos a aplicar el mismo método. En ella hacemos: u = e − x ⇒ du = −e − x dx dv = sen x dx ⇒ v = − cos x Sustituyendo en la expresión anterior nos queda:

∫e

I =e −x sen x +

−x

∫−cos x ⋅( −e

 sen xdx =e −x sen x +−cos x ⋅e −x − 

−x

 )dx  = 

=e −x sen x −cos x ⋅e −x − cos x ⋅e −x dx

es decir, volvemos a la misma integral que pretendemos calcular. Entonces: I = e − x sen x − cos x ⋅ e− x − I

2 I = e − x sen x − cos x ⋅ e − x

⇒ I=

En consecuencia: I=

98.

∫Ln(1 −x ) dx

e − x cos xdx =

e − x (sen x − cos x ) +k 2

Sol: − x − (1 − x ) Ln(1 − x ) + k

−1   −1  u = Ln(1 − x) ⇒ du = dx Ln(1 − x) dx =  1 − x  = xLn(1 − x) − x ⋅ dx = 1− x  dv = dx ⇒ v = x 

19

e − x (sen x − cos x ) 2


INTEGRALES INDEFINIDAS RESUELTAS

= xLn(1 − x ) −

∫ 1 − x dx = xLn(1 − x) − ∫ −x

1 − x −1 dx = xLn(1 − x ) − 1− x

∫ 1 + 1 − x dx = −1 

= xLn(1 − x ) − ( x + Ln(1 − x )) + k = xLn(1 − x ) − x − Ln(1 − x ) + k = = −x − (1 − x ) Ln(1 − x ) + k

99.

x n +1  1   Ln( x ) − +k n +1 n +1

Sol:

x n Ln( x ) dx

Por el método de integración por partes:

100.

1   u = Ln( x ) ⇒ du = dx   1 1 n+ 1 1 n x x Ln( x ) dx =  = x n+ 1 Ln( x ) − x ⋅ dx =  1 n + 1 n + 1 x n n+ 1  dv = x dx ⇒ v = x  n+ 1  

=

1 1 x n +1 Ln( x ) − n +1 n +1

=

x n +1  1   Ln( x ) − +k n +1 n +1

x n dx =

∫arc sen x dx

1 1 1 x n +1 Ln( x ) − ⋅ x n +1 + k = n +1 n +1 n +1

Sol: x arcsen x + 1 − x 2 + k

 1 du = ⋅ dx  u = ar c s e n x   Hacemos el siguiente cambio:  dv = dx  ⇒  1 − x2    v= x Sustituyendo en la fórmula de integración por partes obtenemos:

∫ arc sen x.dx = x.arc sen x − ∫ x ⋅ 1 = x .arc sen x + 2

1 1 − x2

⋅ dx =x .arc sen x −

∫ x.(1 − x ) 2

1

1 2

1 (1 − x 2 ) 2 − 2 x .(1 − x 2 ) .dx = x .arc sen x + ⋅ +k = 1 2 2 −

= x .arc sen x + 1 − x 2 + k

20

1 2

.dx =


INTEGRALES INDEFINIDAS RESUELTAS

101.

∫ 1 −x

2

(

1 − x 2 dx =

1− x2 1− x2

dx =

= arc sen x +

1 1− x2

− x2 1 − x2

dx +

)

1 arcsen x + x 1 − x 2 + k 2

Sol:

dx

− x2 1 − x2

dx =

⋅ dx =

La integral que nos queda la realizaremos por partes:

u = x ⇒ dx   −x −x   ⋅ dx = x ⋅ ⋅ dx =  − x 2 = 1− x2 1 − x 2  dv = 2 dx ⇒ v = 1 − x   1− x 

2

=x

∫ 1 −x

1 −x 2 −

2

dx

Sustituyendo nos queda:

1 − x 2 dx = arc sen x +

− x2 1 − x2

⋅ dx = arc sen x + x 1 − x 2 −

1 − x 2 dx

y se nos repite la misma integral. Entonces:

∫ 1 −x ⇒2

102.

2

∫ 1 −x

dx =arc sen x +x

1 − x 2 dx = arc sen x + x 1 − x 2 ⇒

∫x arc sen x dx

1 −x 2 −

1 − x 2 dx =

Sol:

2

dx

(

)

1 arc sen x + x 1 − x 2 + k 2

[

]

1 ( 2 x 2 − 1) arcsen x + x 1 − x 2 + k 4

 1 du = ⋅ dx  u = arc sen x  1 − x2  Hacemos el siguiente cambio:  ⇒ 2   dv = xdx   x  v = 2 Sustituyendo en la fórmula de integración por partes obtenemos:

21


INTEGRALES INDEFINIDAS RESUELTAS

xarc sen x ⋅ dx =

x2 ⋅ arc sen x − 2

x2 1 x2 1 ⋅ ⋅ dx = ⋅ arc sen x + 2 2 2 2 1− x

− x2 1− x2

dx =

Por el ejercicio anterior tenemos que :

u = x ⇒ dx   −x −x   ⋅ dx = x ⋅ ⋅ dx =  − x 2 = 1− x2 1 − x 2  dv = 2 dx ⇒ v = 1 − x   1− x 

2

= x 1− x − 2

− x2 1 − x2

dx = x 1 − x 2 −

∫ 1 − x dx = x 1− x − ∫

2

1 1 − x2

2

dx −

− x2 1 − x2

1 − x2 1 − x2

dx ⇒

dx = x 1 − x 2 − arc sen x −

− x2 1− x2

En consecuencia:

− x2 1− x2

dx = x 1 − x 2 − arc sen x −

− x2 1 − x2

dx

Por tanto: 2

− x2 1− x

2

dx = x 1 − x 2 − arc sen x ⇒

− x2 1− x

2

dx =

(

1 x 1 − x 2 − arc sen x 2

Sustituyendo obtenemos:

103.

xarc sen x ⋅ dx =

x2 1 ⋅ arc sen x + 2 2

− x2 1 − x2

dx =

(

)

=

x2 1 1 ⋅ arc sen x + ⋅ x 1 − x 2 − arc sen x + k = 2 2 2

=

x2 1 1 ⋅ arc sen x + x 1 − x 2 − arc sen x + k = 2 4 4

=

1 ( 2 x 2 − 1) arcsen x + x 1 − x 2 + k 4

∫arc tg xdx

[

]

1 2

Sol: x arc tg x − Ln (1 + x 2 ) + k

22

)

dx


INTEGRALES INDEFINIDAS RESUELTAS

1   1  u = arc tg x ⇒ du = 2 dx arc tg xdx =  = x ⋅ arc tg x − x ⋅ dx = 1+ x  2 1+ x  dv = dx ⇒ v = x 

= x ⋅ arc tg x −

104.

∫arc tg

x 1 dx = x ⋅ arc tg x − 2 2 1+ x

Sol:

x dx

2x 1 dx = x ⋅ arc tg x − Ln (1 + x 2 ) + k 2 2 1+ x

( x + 1) arc tg

x −

x +k

1 1   1 1  u = arc tg x ⇒ du = ⋅ dx arc tg x dx =  1 + x 2 x  = xarc tg x − x ⋅ dx = 1+ x 2 x   dv = dx ⇒ v = x

1 x 1 t  x = t2 ⇒  = xarc tg x − ⋅ dx =  ⋅ 2tdt =  = xarc tg x − 2 ⇒ dx = 2 tdt 2 1+ x  2 1+ t  x−

t2 dt = xarc tg 1+t2

= xarc tg x −

1+ t2 −1 1   dt = xarc tg x −  1 − dt = 2 2 1+t  1+ t 

= xarc tg

∫ ∫1+ t

105.

∫Ln ( x +

x−

1 + t2 −1 dt = 1+t2

= xarc tg x − dt + = xarc tg

x−

1

x + arc tg

2

dt = xarc tg x − t + arc tg t + k =

x + k = ( x + 1) arc tg

x−

x +k

Sol: x Ln ( x + 1 + x 2 ) − 1 + x 2 + k

1 +x 2 )dx

Hacemos: u = Ln ( x + 1 + x 2 ) y dv = dx con lo cual du =

 2x ⋅  1 + x + 1+ x  2 1 + x2 1

⇒ du =

2

 1  dx =  x + 1 + x2 

 x ⋅  1 + 1 + x2 

 1+ x2 + x  1  dx = ⋅ dx 2  2  x + 1+ x  1+ x 1 + x2  1

Sustituyendo en la fórmula de integración por partes, obtenemos:

∫ Ln ( x +

1 + x 2 )dx = x ⋅ Ln ( x + 1 + x 2 ) − 23

∫x⋅

1 1 + x2

dx =

y

  dx ⇒  

v= x


INTEGRALES INDEFINIDAS RESUELTAS

= x ⋅ Ln ( x + 1 + x 2 ) −

106.

x arc sen x 1 −x2

∫ 2 1+ x 2x

dx

2

dx = x ⋅ Ln ( x + 1 + x 2 ) − 1 + x 2 + k

Sol: x − 1 − x 2 arc sen x + k

1   u = arc sen x ⇒ du = dx  2 x arc sen x  1 − x dx =  = 2 x − 2 x 2 1− x  dv = dx ⇒ v = − dx = − 1 − x  2 2  1 − x  2 1− x

= − 1 − x 2 ⋅ arc sen x −

− 1 − x2 ⋅

1 1− x

2

dx = − 1 − x 2 ⋅ arc sen x +

∫ dx =

= − 1 − x 2 ⋅ arc sen x + x + k

107.

( x − 2)3 +k Sol: Ln x −1

2 x −1 dx ( x − 1)( x − 2)

Tenemos una integral de tipo racional donde el grado del numerador es menor que el grado del denominador. Vamos a descomponer el integrando en fracciones simples:

 x = 1 (raíces reales simples) ( x − 1)( x − 2) = 0 ⇒   x= 2 Entonces: 2x −1 A B A( x − 2) + B ( x − 1) = + = ( x − 1)( x − 2) x − 1 x − 2 ( x − 1)( x − 2)

Vamos a calcular los coeficientes indeterminados. Al ser los denominadores iguales, los numeradores también lo serán. Por tanto:

 x = 1→ 1= − A⇒ A= − 1 2x − 1 = A(x − 2) + B(x − 1) ⇒   x= 2→ 3= B ⇒ B= 3 Por tanto,

24


INTEGRALES INDEFINIDAS RESUELTAS

∫ ( x − 1)( x − 2) dx = ∫  x − 1 + x − 2  dx = −∫ x − 1 dx + 3∫ x − 2 dx = 2x −1

 −1

3

1

= −Ln ( x − 1) + 3 Ln ( x − 2) + k = Ln

108.

xdx ( x + 1)( x + 3)( x + 5)

1

( x − 2)3 +k x −1

1 ( x + 3)6 Ln +k 8 ( x + 1)( x + 5)5

Sol:

Tenemos una integral de tipo racional donde el grado del numerador es menor que el grado del denominador. Vamos a descomponer el integrando en fracciones simples:

x= −1  ( x + 1)( x + 3)( x + 5) = 0 ⇒  x = − 3  x = − 5

(raíces reales simples)

Entonces: x A B C = + + = ( x + 1)( x + 3)( x + 5 x + 1 x + 3 x + 5

=

A( x + 3)( x + 5) + B( x + 1)( x + 5) + C ( x + 1)( x + 3) ( x + 1)( x + 3)( x + 5)

Para calcular los coeficientes indeterminados, al ser los denominadores iguales, los numeradores también lo serán. Por tanto:

1  x = − 1 → − 1 = 8 A ⇒ A = − 8  3 x = A( x + 3)( x + 5) + B( x + 1)( x + 5) + C ( x + 1)( x + 3) ⇒  x = − 3 → − 3 = − 4B ⇒ B = 4   x = − 5 → − 5 = 8C ⇒ C = − 5  8 Por tanto,

∫ =−

1 8

1 3 dx + x +1 4

xdx = ( x +1)( x + 3)( x + 5)

1 5 dx − x +3 8

1 5  3  −  −  8 8 dx = 4  + +  x +1 x + 3 x +5     

1 1 3 5 dx = − Ln( x + 1) + Ln( x + 3) − Ln( x + 5) + k = x +5 8 4 8 25


INTEGRALES INDEFINIDAS RESUELTAS

=

109.

6 1 ( Ln ( x + 1) + 6 Ln ( x + 3) − 5 Ln ( x + 5)) + k = 1 Ln ( x + 3) 5 + k 8 8 ( x + 1)( x + 5)

x3 x 2 x 2 ( x − 2)5 + + 4 x + Ln +k Sol: 3 2 ( x + 2)3

x5 + x4 − 8 dx x3 − 4x

Al ser el grado del numerador mayor que el grado del denominador, antes de aplicar el método de descomposición en fracciones simples tendremos que dividir. De esta forma obtenemos: x5 + x 4 − 8 4 x 2 + 16 x − 8 2 = x + x + 4 + x3 − 4 x x3 − 4 x En consecuencia:

x5 + x 4 − 8 ⋅ dx = x3 − 4 x =

( x 2 + x + 4)dx +

x3 x 2 + + 4x + 3 2

4 x 2 + 16 x − 8 ⋅ dx = x3 − 4 x

4 x 2 + 16 x − 8 ⋅ dx x3 − 4 x

A la integral que nos queda le aplicamos el método de descomposición en fracciones simples. Calculamos las raíces del denominador: x = 0 x 3 − 4 x = 0 → x ⋅ ( x 2 − 4) = 0 →   x = ±2 Entonces: 4 x 2 + 16 x − 8 A B C A( x − 2)( x + 2) + Bx ( x + 2) + Cx ( x − 2) = + + = 3 x − 4x x x−2 x+2 x ( x − 2)( x + 2) Como los denominadores son iguales, los numeradores también lo serán; por tanto: 4 x 2 + 16 x − 8 = A( x − 2)( x + 2) + Bx( x + 2) + Cx( x − 2) Calculamos los coeficientes indeterminados: le vamos asignando los valores de las raíces x=0 → − 8 = −4 A → A = 2 x=2 → 40 = 8B → B = 5 x = −2 → − 24 = 8C → C = −3 Por tanto, la fracción descompuesta en fracciones simples nos queda:

26


INTEGRALES INDEFINIDAS RESUELTAS

4 x 2 + 16 x − 8 2 5 3 = + − 3 x − 4x x x−2 x+2 La integral de la función pedida será:

x5 + x 4 − 8 x3 x 2 ⋅ dx = + + 4x + x3 − 4 x 3 2 =

4 x 2 + 16 x − 8 ⋅ dx = x3 − 4 x

x3 x 2 5 3  2 + + 4x +  + −  ⋅ dx 3 2  x x−2 x+2

x3 x 2 = + + 4x + 3 2

2 ⋅ dx + x

5 ⋅ dx − x−2

3 ⋅ dx = x+2

=

x3 x 2 1 1 1 + + 4x + 2 ⋅ dx + 5 ⋅ dx − 3 ⋅ dx = 3 2 x x−2 x+2

=

x3 x 2 + + 4 x + 2 ⋅ Ln | x | +5 ⋅ Ln | x − 2 | −3 ⋅ Ln | x + 2 | + k = 3 2

x3 x 2 x 2 ( x − 2)5 = + + 4 x + Ln +k 3 2 ( x + 2)3 110.

x 4 dx ( x 2 −1)( x + 2)

Sol:

x2 1 ( x − 1) 16 − 2x + Ln + ⋅ Ln | x + 2 | + k 3 2 6 ( x + 1) 3

Como el grado del numerador es mayor que el del denominador, tenemos que dividir, obteniendo: x4 5x2 − 4 = x − 2 + ( x 2 − 1)( x + 2) ( x 2 − 1)( x + 2) Con lo que

x 4 ⋅ dx = ( x 2 − 1)( x + 2)

( x − 2) dx +

5x2 − 4 x2 dx = − 2x + ( x 2 − 1)( x + 2) 2

5x2 − 4 dx ( x 2 − 1)( x + 2)

y tendremos que integrar la función racional que nos queda, donde el grado del numerador es menor que el grado del denominador.

Descomponemos en fracciones simples: 27


INTEGRALES INDEFINIDAS RESUELTAS

5x2 − 4 A B C A( x + 1)( x + 2) + B ( x − 1)( x + 2) + C ( x − 1)( x + 1) = + + = 2 ( x − 1)( x + 2) x − 1 x + 1 x + 2 ( x 2 − 1)( x + 2) Como los denominadores son iguales, también lo serán los numeradores. Entonces: 5 x 2 − 4 = A( x + 1)( x + 2) + B( x − 1)( x + 2) + C ( x − 1)( x + 1) Calculamos los coeficientes indeterminados: x =1

→ 1 = 6A

A=

1 6

1 2 16 x = −2 → 16 = 3C → C = 3 x = −1 → 1 = − 2 B → B = −

1 16 1 − 5x2 − 4 = 6 + 2 + 3 2 ( x − 1)( x + 2) x − 1 x + 1 x + 2

Entonces:

Y, por tanto:

x ⋅ dx x = − 2x + ( x − 1)( x + 2) 2 4

2

2

x2 = − 2x + 2

111.

∫ ( x −1) ( x − 2) dx 2

∫ ∫

16  1  1 −  5x − 4 x 6 + 2 + 3  ⋅ dx = dx = − 2 x +   ( x 2 − 1)( x + 2) 2  x − 1 x + 1 x + 2    2

1 6 ⋅ dx + x −1

2

1 2 ⋅ dx + x +1 −

16 3 ⋅ dx = x+2

=

x2 1 − 2x + 2 6

=

x2 1 1 16 − 2 x + ⋅ Ln | x − 1| − ⋅ Ln | x + 1| + ⋅ Ln | x + 2 | + k = 2 6 2 3

=

x2 1 16 − 2 x + ⋅ ( Ln | x − 1| −3 ⋅ Ln | x + 1| ) + ⋅ Ln | x + 2 | + k = 2 6 3

=

x2 1 x −1 16 − 2 x + ⋅ Ln + ⋅ Ln | x + 2 | + k 3 2 6 ( x + 1) 3

1 1 ⋅ dx − x −1 2

Sol: 28

1 16 ⋅ dx + x +1 3

1 ⋅ dx = x+2

1 x−2 + Ln +k x −1 x −1


INTEGRALES INDEFINIDAS RESUELTAS

Como el grado del numerador es menor que el grado del denominador aplicamos la descomposición en fracciones simples directamente: 1 A B C A( x − 1)( x − 2) + B ( x − 2) + C ( x − 1) 2 = + + = ( x − 1) 2 ( x − 2) ( x − 1) ( x − 1) 2 ( x − 2) ( x − 1) 2 ( x − 2) → 1 = A( x − 1)( x − 2) + B ( x − 2) + C ( x − 1) 2 Calculamos los coeficientes: x = 1 : 1 = − B → B = −1 x = 2: 1= C x = 0 : 1 = 2 A − 2B + C → 1 = 2 A + 2 + 1 →

A = −1

Entonces:

1 ⋅ dx = ( x − 1) ( x − 2) 2

=−

−1 ⋅ dx + x −1

−1 ⋅ dx + ( x − 1) 2

1 ⋅ dx − ( x − 1) −2 ⋅ dx + x −1

1 ⋅ dx = ( x − 2)

1 ⋅ dx = x−2

( x − 1) −1 1 x−2 = − Ln | x − 1| − + Ln | x − 2 | + k = + Ln +k −1 x −1 x −1

112.

x −8 dx 3 x − 4x 2 + 4x

3 ( x − 2) 2 Sol: + Ln +k x−2 x2

Igual que en el anterior, aplicamos la descomposición en fracciones simples: Calculamos las raíces del denominador: x = 0 x 3 − 4 x 2 + 4 x = 0 → x ⋅ ( x 2 − 4 x + 4) = 0 → x ⋅ ( x − 2) 2 = 0 →   x = 2 (doble) Entonces: x −8 A B C A( x − 2) 2 + Bx ( x − 2) + Cx = + + = x 3 − 4 x 2 + 4 x x x − 2 ( x − 2) 2 x ( x − 2) 2 ⇒

x − 8 = A( x − 2) 2 + Bx( x − 2) + Cx ⇒

Calculamos los coeficientes:

29


INTEGRALES INDEFINIDAS RESUELTAS

x = 0 → − 8 = 4 A → A = −2 x = 2 → − 6 = 2C → C = −3 x =1 → −7 = A− B +C → B = 7 + A+C = 7 − 2−3 = 2 → B = 2 Entonces:

x −8 dx = x − 4x2 + 4x 3

= −2

 −2 2 −3  + ⋅ dx =  + 2   x x − 2 ( x − 2) 

= −2 Ln | x | +2 Ln | x − 2 | −3 ⋅

113.

114.

1 1 1 ⋅ dx + 2 ⋅ dx − 3 ⋅ dx = −2 Ln | x | +2 Ln | x − 2 | −3 ( x − 2) −2 dx = x x−2 ( x − 2) 2 ( x − 2) −1 3 ( x − 2) 2 +k = + Ln +k −1 x−2 x2

3x + 2 dx x( x + 1)3

Sol:

3x − 2 dx x ( x +1) 3

Sol: Ln

4x + 3 x2 + Ln +k 2( x + 1) 2 ( x + 1) 2

( x + 1) 2 4x + 9 − +k 2 x 2( x + 1) 2

Descomponemos el integrando en fracciones simples: 3x − 2 A B C D A( x + 1)3 + Bx ( x + 1) 2 + Cx( x + 1) + Dx = + + + = x( x + 1)3 x x + 1 ( x + 1) 2 ( x + 1)3 x ⋅ ( x + 1)3

→ 3 x − 2 = A( x + 1)3 + Bx( x + 1) 2 + Cx ( x + 1) + Dx → Calculamos los coeficientes: x=0 → x = −1 → x =1 → x = −2 →

−2= A − 5 = −D → D = 5 1 = 8 A + 4 B + 2C + D → 1 = −16 + 4 B + 2C + 5 → 2 B + C = 6 − 8 = − A − 2 B + 2C − 2 D → − 8 = 2 − 2 B + 2C − 10 → − B + C = 0

Resolviendo el sistema resultante, obtenemos: 2B + C = 6  2 B + B = 6 → B = 2  →  − B + C = 0 C = B = 2 Entonces:

3x − 2 ⋅ dx = x( x + 1)3

−2 ⋅ dx + x

2 ⋅ dx + x +1

∫ 30

2 ⋅ dx + ( x + 1) 2

5 ⋅ dx = ( x + 1)3


INTEGRALES INDEFINIDAS RESUELTAS

= −2

1 ⋅ dx + 2 x

1 ⋅ dx + 2 x +1

= −2 ⋅ Ln | x | +2 ⋅ Ln | x + 1| +2

1 ⋅ dx + 5 ( x + 1) 2

( x + 1) −2 ⋅ dx + 5

1 ⋅ dx = ( x + 1)3

( x + 1) −3 ⋅ dx =

( x + 1) −1 ( x + 1)−2 = −2 ⋅ Ln | x | +2 ⋅ Ln | x + 1| +2 +5 +k = −1 −2 = Ln

( x + 1) 2 2 5 ( x + 1) 2 4( x + 1) + 5 − − + k = Ln − +k = x2 x + 1 2( x + 1) 2 x2 2( x + 1) 2

= Ln

( x + 1) 2 4x + 9 − +k 2 x 2( x + 1) 2

115.

116.

117.

x3 − 3 x dx 64 x

6

118.

119.

120.

121.

∫e

122.

x 2 dx ( x + 2) 2 ( x + 4) 2

x 4

6

x +1 3

x + x 7

7

dx

x +1

7

4

5

x +

14

x

15

dx e +1 x

4 3

Sol:

2 4 9 2 x − 12 x 13 + k 27 13

(

4

x 3 − Ln

(

4

))

x3 +1 + k

6 12 + 12 + 2 Ln x − 24 Ln (12 x + 1) + k x x 1

1

1

1

 1− x 3 1− x 6  − + 1 − x − Ln 1 + 6 1 − x  + k 2  3 

dx 1− x + 3 1− x

2x

Sol:

14 3 x − 7 x 2 + 14 x 5  + k Sol: 4 14 x − 7 x + 2 3 4 5  

dx

ex dx + ex − 2

5 x + 12  x+4 + Ln   +k 2 x + 6x + 8  x+2

Sol: − 6

dx

x+ x 8

2

Sol: −

Sol: 6

Sol:

1 ex −1 Ln x +k 3 e +2 Sol: x − Ln (e x +1) + k

31

(

)


INTEGRALES INDEFINIDAS RESUELTAS

123.

∫e

124.

∫sen

2x

ex dx + 3e x + 2 3

x dx

Sol: Ln

Sol:

125.

32

ex + 1 +k ex + 2

1 cos 3 x − cos x + k 3

Integrales indefinidas resueltas 02  
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