Integrales indefinidas

1. Hallar: ∫ 1

x2+4dx

Solución:

Se sabe que: d ( 12arctg

x2 )= 1

x2+4dx

→∫ 1

x2+4dx=∫ d (1

2arctg

x2 )=1

2arctg

x2+C

∴∫ 1

x2+4dx=1

2arctg

x2+C

2. Hallar: ∫ x2+2x2(x2+4)

dx

Solución:

∫ x2+2

x2 (x2+4 )dx=∫ x2+4−2

x2(x2+4)dx=∫ [ 1

x2−2

x2(x2+4) ]dx∫ x−2dx−∫ 2

x2 (x2+4 )dx=∫ x−2dx−∫ x2+4−x2

2 x2(x2+4)dx

∫ x−2dx−[∫ x−2

2dx−∫ dx

2 (x2+4 ) ]=∫ x−2dx−∫ x−2

2dx+∫ dx

2 (x2+4 )

x−2+1

−2+1−1

2 ( x−2+1

−2+1 )+ 12 ( 1

2arctan

x2 )+C

−12 x

+ 14

arctanx2+C

3. Hallar: ∫ (x3+1 )43 x2dx

Solución:

Sea: u=x3+1 ;u=u ( x )→fd

du=d (x3+1 )=(x3+1 )' dx=3 x2dx→∴du=3 x2dx

∫(x3+1)4 3 x2dx=∫ u4du=u5

5+c

∴∫ (x3+1 )43 x2dx=

(x3+1)5

5+c

4. Hallar: ∫ x4

7√ x5+1

Solución:

x5+1→d (x5+1 )=(x5+1 )' dx=5 x4dx

→∫ d (x5+1)

57√ x5+1

=15∫ d (x5+1)

7√ x5+1=

15

(x5+1)−17

+1

−17

+1

∴∫ x4

7√ x5+1= 7

30(x5+1)

67 +C

5. Hallar: ∫ 5 ex

√1−e2xdx

Solución:

∫ 5 ex

√1−e2xdx=5∫ 1

√1−(ex)2dex

∴∫ 5e x

√1−e2xdx=5 arcsen (ex )+C

6. Hallar: ∫ arcsen√ x√x−x2

dx

Solución:

Sea: u¿arcsen√ x

d (arcsen√ x )= dx

2√x−x2

→∫ arcsen √x√x−x2

dx=2∫arcsen√ x .d (arcsen√ x )

∴∫ arcsen√ x√x−x2

dx=arcsen2√ x+C

7. Hallar: ∫ √2+√2+√2+2 cos(5√ x+4) x−12 dx

Solución:

Tenemos: 2+2cos (5√ x+4 )=2¿

→∫ x−12 √2+√2+√4cos2 5√x+4

2dx=∫ √2+√2+2cos (5 √x+4

2 ) x−12 dx

→∫ √2+√2+√2+2 cos(5√x+4 )x−12 dx=∫ 2 cos( 5√ x+4

8 )x−1

2 dx

→325∫cos ( 5√x+4

8 )d ( 5√ x+48 )

∴∫√2+√2+√2+2cos (5√ x+4) x−12 d x=32

5sen ( 5√x+4

8 )+C

8. Hallar: ∫ x

e3x (1−x )4dx

Solución:

∫ e x . xex . e3 x(1−x)4 =∫ ex . x

e4 x (1−x)4 dx

Tomaremos: ex−x ex=z→dz=[ex−(ex+ xex ) ]dx=−ex dx

∫−dzz4 =(−1 )∫ z−4dz=−z−4+1

−4+1+C=1

3z−3+C= 1

3 z3 +C= 13(ex−x ex )3

+C

9. Hallar: ∫ x2−1

(x¿¿2+1)√x4+1dx¿

Solución:

∫ x2−1

(x¿¿2+1)√x4+1dx=∫

x2(1− 1

x2 )x (1+ 1

x )√x2(1+ 1x2 )

dx ¿

d (x+ 1x )=[1+(−1)( x−2)]dx=(1− 1

x2 )dxz=(x+ 1

x )

z2=x2+ 1

x2+2→x2+ 1

x2=z2−2

→∫(1− 1

x2 )(x+ 1

x )√(x+ 1x )

2

−2

dx=∫d (x+ 1

x )( x+ 1

x )√(x+ 1x )

2

−2

=∫ d (z )

z √z2−√22

∴∫ x2−1

(x¿¿2+1)√x4+1dx=

1

√2arcsec ( x+ 1

x

√2 )+C ¿

10. Hallar: ∫ x+2

( x−2 )4dx

Solución:

Sabemos: d ( x−2 )=dx

∫ x+2

( x−2 )4d (x−2 )=∫ x−2+4

( x−2 )4d ( x−2 )=∫ (x−2)d ( x−2 )

( x−2 ) 4 +4∫ 1

( x−2 )4d ( x−2 )=∫ d ( x−2 )

( x−2 )3+4∫ 1

( x−2 )4d ( x−2 )

∴∫ ( x+2 )( x−2 )4

dx=−43

1

( x−2 )3−1

21

( x−2 )2+C

11. Hallar: ∫ x

√ x2+1√1+√x2+1dx

Solución:

Sabemos: 1+√ x2+1→d (1+√x2+1 )=(1+√ x2+1 ) 'dx= x

√ x2+1dx

∫ d (1+√ x2+1)

√1+√x2+1=∫(1+√x2+1)

−12 d (1+√ x2+1 ) ;u=1+√ x2+1

∫u−12 d (u )=2u

12 +C∴∫ x

√ x2+1√1+√x2+1dx=2√1+√ x2+1+C

12. Hallar: ∫ x √x+4dx

Solución:

∫ x √x+4d ( x+4 )=∫ ( x+4−4 ) √x+4d ( x+4 )=∫ ( x+4 )√ x+4 d (x+4)−∫ 4√ x+4d (x+4)=∫ ( x+4 )3 /2d (x+4)−4∫√ x+4 d (x+4)

∴∫ x √x+4dx=2(x+4 )

52

5−

8(x+4)32

3+C

13. Hallar: ∫ x2 e3 xdx

Solución:

∫ x2 e3 xdx=(A x2+Bx+C )e3x

x2 e3 x=(2 Ax+B ) e3x+3(e3 x)( A x2+Bx+C )x2 e3 x=2 Ae3x x+B .e3x+3 A e3 x x2+3Be3x x+3Ce3x

x2 e3 x=3 A e3 x x2+ (2 A+3 B ) e3x x+B .e3 x+3C e3 x

x2=3 A x2+(2 A+3 B ) x+ (B+3C )

A=13;B=−2

9;C= 2

27

∴∫ x2 e3 xdx=¿( x2

3−2 x

9+ 2

27 )e3x¿

14. Hallar: ∫ (x2+3 x−1 )e2x dx

Solución:

∫ (x2+3 x−1 )e2x dx=Pm ( x )e2x+C

d (∫ (x2+3 x−1 )e2x dx)=d [ ( A x2+Bx+C )e2 x+C ](x2+3 x−1 )e2 x=(2 Ax+B )e2x+2e2x (A x2+Bx+C )

x2+3x−1=(2 A ) x2+2 (A+B ) x+(B+2C )

A=12;B=1;C=−1

∴∫ (x2+3 x−1 )e2x dx=( x2

2+x−1)e2 x+C

15. Hallar: ∫ sec3(x )dx

Solución:

∫ sec x . sec2dx=∫ sec x .d (tgx )=∫ ( sec2 )12 . d (tgx)

∫ (1+ tg2 x )12d (tgx )=∫ (1+u2 )

12d (u ) ;donde→u=tgx

∴∫ sec3(x)dx=12

[tgx √tg2 x+1+ ln (tg+√ tg2+1)]+C16. Hallar: ∫ sec5 x dx

Solución:

∫ sec3 x . sec2 xdx→u=sec3 x¿ x dx=d (v )=d (tg)¿

¿

∫ sec5 x dx=sec3 x tgx−∫ tg ( x )d (sec 3 x )donde : d (sec 3 x)=3 . sec2 x . sec x .tgx . dx

d ( sec3 x )=3 sec3 x . tgx .dx

∫ sec5 x dx=sec3 x tgx−∫ tgx (3 sec3 . tgx .dx )

∫ sec5 x dx=sec3 x tgx−∫ tg2 x .3 sec 3 x dx

∫ sec5 x dx=sec3 x tgx−∫ (sec2 x−1 ) 3 sec3 x d x

∫ sec5 x dx=sec3 x tgx−∫ (3 sec5 x−3 sec3 x )dx

∫ sec5 x dx=sec3 x tgx−3∫ sec5 xdx+3∫sec 3 x dx

∫ sec5 x dx=sec3 x tgx+3∫ sec3 x dx

4

∴∫ sec5 x dx=sec 3 x .tgx+3[ 1

2( tgx√ tg2 x+1+ln (tgx+√tg2 x+1 ))]

4+C

17. Hallar: ∫ x arctgx dx

Solución:

∫ x arctgx dx=uv−∫ vduDonde :u=arctg ( x ); x=d (v )

∫ x arctgx dx= x2

2arctgx−∫ x2

2d (arctgx )

∫ x arctgx dx= x2

2arctgx−∫ x2

2.

dx(x2+1 )

=12 (x2arctgx−∫ x2+1−1

(x2+1 )dx )=1

2 (x2arctgx−∫ dx+∫ 1(x2+1 )

dx )∴∫ x arctgx dx=1

2(x2arctgx−x+arctgx )+C=1

2[ (x2+1 )arctgx−x ]+C

18. Hallar: ∫ cosx+x . senx−1

( senx−x )2dx

Solución:

∫ cosx+xsenx−1( senx−x )2

dx=∫ cosx+xsenx−sen2 x−cos2 x( senx−x )2

dx=−∫ senx( senx−x )

dx−∫ cosx (cosx−1 )(senx−x )2

dx=−∫ senx( senx−x )

dx−[uv−∫ vdu ]

Donde :u=cosx;−(cosx−1)(senx−x )2

dx=d ( 1senx−x )=dv

→∫ cosx+xsenx−1

( senx−x )2dx=−∫ senx

(senx−x )dx−[cosx 1

senx−x−∫ senx

senx−xdx]

∴∫ cosx+xsenx−1

( senx−x )2dx= −cosx

senx−x+C

19. Hallar: ∫ ex

x(1+xlnx )dx

Solución:

∫ ex

x(1+xlnx )dx=∫ e x

xdx+∫ ex . lnx .dx∫ ex

x=uv−∫ vdu;Donde :u=ex ;

1xdx=dv

∫ ex

x(1+xlnx )dx=ex . lnx−∫ lnx .d (e x)+∫ ex . lnx . dx=ex .lnx−∫ex .lnx d ( x )+∫ ex . lnx . d (x)

∴∫ ex

x(1+xlnx )dx=ex . lnx+C

20. Hallar: ∫xearctgx

(1+x2)32

dx

Solución:

∫ x earctgx

(1+x2 )32

dx=∫( x2

x2+1 )12 e

arctgx

x2+1dx=uv−∫ vdu

Donde: u=( x2

x2+1)1 /2

;dv= earctgx

x2+1dx=d (e¿¿arctgx)¿

→∫ x earctgx

(1+x2 )32

dx=√ x2

x2+1earctgx−∫ earctgx √ x2+1

(x2+1 )2dx=√ x2

x2+1earctgx−¿

Donde:u1=1

√ x2+1;d (v1 )=d (earctgx )

→∫ x earctgx

(1+x2 )32

dx= x

√ x2+1earctgx−¿

→∫ x earctgx

(1+ x2)32

dx=

x

√x2+1earctgx− 1

√x2+1earctgx

2

∴∫ x earctgx √x2+1

( x2+1 )2dx=√ x2+1earctgx (x−1)

(x¿¿2+1).2+C ¿

21. Hallar: ∫ esenx (x cos3 x−senx )cos2 x

dx

Solución:

∫(esenx . x . cosx−esenx

.senx

cos2 x )dx=∫ x (esenx.cosx )dx−∫ (esenx )tg . x . secx .dx

¿ (uv−∫ vdu )−(u1 v1−∫ v1d u1 )Donde: u=x ; (esenx . cosx )dx=dv ;esenx=u1; tg . x . secx .dx=d v2

→∫ esenx (x cos3 x−senx )cos2 x

dx=x esenx−∫esenx dx−esenx secx+∫sec d (esenx )=x esenx−∫ esenxdx−esenx . secx+∫esenxdx

∴∫ esenx (x cos3 x−senx )cos2 x

dx=esenx ( x−secx )+C

22. Hallar: ∫ sen3 x dx

Solución:

∫ sen2 x . senx.dx=∫ (1−cos2 x ) senx .dx

∫ senx .dx−∫cos2 x senx dx=−∫ d (cosx )+∫ cos2 xd (cosx)

∴∫ sen3 x dx=−cosx+¿ cos3 x3

+C ¿

23. Hallar: ∫ sen3 x .cos2 x .dx

Solución:

∫ sen3 x .cos2 x .dx=∫senx . sen2 .cos2 x .dx=−∫ (1−cos2 x ) cos2 x .d (cosx )=−∫ ( cos2 x−cos4 x )d (cosx )=−∫ cos2 x d (cosx )+∫ cos4 x .d (cosx)

∴∫ sen3 x .cos2 x .dx=−cos3 x3

+ cos5 x5

+C

24. Hallar: ∫cos2 3x dx

Solución:12∫ (1+cos6 x )dx=1

2∫d (x )+ 12∫ cos6 x .dx=1

2∫d ( x )+ 112∫ cos6 x .d 6 x

∴∫cos23 x dx=12x+ 1

12sen6 x+C

25. Hallar: ∫ sen4 x .dx

Solucion:

∫ sen4 x .dx=∫ (1−cos2 x )2dx¿∫ (cos4 x−2cos2 x+1 )dx

¿∫cos4 xdx−∫cos2 xdx¿∫ (1+cos2x )2

4−1

2∫cos2 x .d 2x

¿∫( cos22 x4

+ cos2 x2

+ 14 )dx−1

2∫cos2 x .d2 x

¿∫ cos2 2x4

dx+∫ cos2 x2

dx+ 14∫dx−1

2∫cos 2x .d2 x

¿ 18∫ (1+cos 4 x )dx+∫ cos 2x

4d 2 x+ 1

4∫dx−12∫ cos2x .d2 x

¿ 18∫dx+∫ cos 4 x

32d 4 x+∫ cos2x

4d2 x+ 1

4∫ dx−12∫ cos2 x .d 2x

¿∫ cos4 x32

d 4 x−14∫cos2 x .d 2x+ 3

8dx

∴∫ sen4 x .dx= sen 4 x32

− sen2 x4

+ 38x+C

26. Hallar: ∫ sen2 2 x .cos4 .2 x .dx

Solucion:

∫ sen2 2 x .cos4 .2 x .dx=∫ (sen2 x .cos 2x )2 .cos22 x .dx

¿ 14∫ sen2 4 x .cos22 x .dx

¿ 14∫ sen2 4 x ( 1+cos4 x

2 ). dx

¿ 18∫ ( sen2 4 x+sen2 4 x .cos4 x )dx

¿ 18∫( 1−cos8 x

2+sen2 4 x cos4 x)dx

¿ 116∫(1−cos8 x+2 sen¿¿2 4 xcos 4 x )dx ¿

¿ 116∫dx− 1

128∫ cos8 x .d 8 x+ 132∫ sen2 4 x .dsen4 x

∴∫ sen2 xcos4 2 x¿ . dx= 116

x−sen (8 x )

128+ 1

96sen3 4 x¿+C

27. Hallar: ∫ tg3

Sec 4 xx .dx

Solución:

∫ tg3 x . sec xsec 4 . secx .

dx=∫ tg2 x . tgx . secxsec4 x . secx

.dx

¿∫ tg2 xsec5 x

d (sec¿x)¿

¿∫( sec2

sec5 −1

sec5 )d (sec )

¿∫ d ( secx )sec3 x

−∫ d (sec x )sec5 x

∴ tg3 . dxsec 4 x

= 14 sec4 x

− 12 sec2 x

+C

28. Hallar: ∫ ctg5 x .dx

Solución:

∫ ctg4 x . ctg x .dx=∫ctg 4 x . ctgcsecxcsecx

. dx

¿−∫ ctg4 xcsecx

(−ctgx)(csecx)dx

¿−∫ ctg4 xcsecx

.d (cscx )

¿−∫ (csec2 x−1)2

csecx.dx

¿−∫( csec 4 xcsecx

−2csec2 xcsec x

+ 1csecx )d (csecx)

¿−∫(csec3 x−2csecx+ 1csecx )d (csecx)

¿−∫ csec3 xd (csecx )+2∫csecx d (csecx )−∫ 1csecx

d (csecx)

∴∫ ct g5 x .dx=−csec 4 x4

+csec2 x−ln|csecx|+C

29. Hallar: ∫ tg1 /2 x sec4 xdx

Solución:

∫ tg12 sec2 x . d (tgx )=∫ tg

12 x (1+tg2 x )d (tgx )

¿ tg1/2 x d (tgx )+∫ tg5/2 x d ( tgx )

∴∫ tg1 /2 x sec4 xdx=( tg3/2 x ) 23+

2(tg72)

7+C

30. Hallar: ∫ csec4 3 xdx

Solución

∫ csc23 x . csec 23 x dx=∫ (1+ctg23 x )csec2 3 xdx

¿ 1−3∫ (1+ctg2 3 x ) (−3 (csec2 3x )) dx

¿−13∫ (1+ctg23 x )d (ctg3 x )

¿ −13 ∫ d (ctg3 x )−1

3∫ ctg23 xd (ctg 3 x )

∴∫ csc4 3x dx=−13

ctg 3 x−19ctg3 3 x+C

31. Hallar: ∫ sen2 xcos3 x

Solución:

∫ sen2 xcos3 x=12∫−2 sen2x cos3 x

¿−12∫ (senx−sen5 x )dx

¿−12

(∫ senxdx−∫ sen5 xdx )

¿−12 (∫ senxdx−1

5∫ sen5 xd5 x)

∴∫ sen2 xcos3 x=12cosx− 1

10cos5 x+C

32. Hallar: ∫ sen3 3 x . tg3 x .dx

Solución:

∫ sen4 3 xcos3 x

.dx=13∫ sen4 3x

cos23 xd (sen 3x )

¿ 13∫ (sen¿¿ 43 x−1+1)

1−sen2 .3 xd ( sen3 x )¿

¿ 13 [−∫ (1−sen23 x )(1+sen2 3 x)

(1−sen2 3x )d (sen 3x )+∫ d ( sen3 x )

1−sen2 3x ]13 [−∫ d ( sen3 x )−∫ sen23 x d (sen 3x )−∫ 1

1−sen23 xd ( sen3 x )]

∴∫ sen3 3 x . tg3 x .dx=13 [−sen3 x− sen3 3 x

3−1

2ln( 1+sen 3x

1−sen3 x )]+C33. Hallar: ∫cos 7 x .cos3 x dx

Solución:

12∫ 2 cos7 x .cos3 x=1

2∫¿¿¿¿¿

12∫ cos10 x dx+ 1

2∫cos 4 x dx= 120∫cos10 x d (10 x )+ 1

8∫cos 4 x d (4 x)

∴∫cos 7 x .cos3 x dx= 120

sen (10x )+ 18sen (4 x )+C

34. Hallar: ∫ sen3 x . sen2x dx

Solución:

∫ sen3 x . sen2x dx=−12 ∫−2 sen 3x . sen2 x dx

¿−12∫ (cos5 x−cosx )dx

¿−12 [ 1

5∫ (cos5 x )d5 x−∫ (cosx )dx ]

∴∫ sen3 x . sen2x dx=−110

sen (5x )+ 12

cos (x )+C

35. Hallar : ∫ sen4 x+cos4 xsen2 x−cos2 x

dx

Solución:

∫ sen4 x+cos4 x

sen2 x−cos2 xdx=∫ ( sen2 x+cos2 x )2−2 sen2 x .cos2 x

sen2 x−cos2 x.dx

¿−∫1−

(2 senx . cosx )2

2cos 2x

=−∫1− sen2 2 x

2cos 2x

.dx

−∫ 1cos2 x

dx+∫ sen22 x2 cos2 x

dx=−∫ sec 2x .dx+∫ (1−cos22 x)2 cos2 x

dx

¿−∫ sec 2 x .dx+12∫ sec 2 xdx−1

2∫cos 2x .dx

X

1

¿−14∫ sec 2x .d2 x−1

4∫cos2 x .d 2 x

∴∫ sen4 x+cos4 xsen2 x−cos2 x

dx=−14

[ ln|tg ( sec2 x+tg2 x )|]−14sen2 x+C

36. Hallar: ∫ sen3 3 x . tg3 x .dx

Solución:

∫ sen3 3 x . tg3 x .dx=∫ sen43 xcos 3x

dx=∫ ( sen23 x )2

cos3 xdx=∫ (1−cos2 3x )2

cos3 xdx

¿∫( 1cos 3x

−2cos3 x+cos3 3 x)dx

¿∫ 1cos3 x

dx−2∫ cos3 xdx+∫ cos2 3x .dsen3 x

3

¿ 13∫ sec3 x .d 3 x−2

3∫cos3 xd 3x+∫ (1−sen23 x )dsen3 x

¿ 13∫ sec3 x .d 3 x−2

3∫cos3 xd 3x+ 13∫dsen 3x−1

3∫ sen23 x .dsen3 x

∴∫ sen3 3 x . tg3 x .dx=[ ln|tg ( sec 3x+tg 2x )|]

3−

13sen3 x−

sen33 x9

+C

37. Hallar : ∫ √ x2−1x

.dx

Solución:

X

1

Sustituyendo: ∫ √ x2−1dxx

=∫ senθ secθ tgθdθ

¿∫ tg2θdθ=∫ ( sec2−1 )dθ

¿∫ sec2θdθ−∫dθ=tangθ−θ+C

∴∫ √ x2−1x

.dx=√ x2−1−arccos ( x )+C

38. Hallar: ∫ √ x2+1x

dx

Solucion:

Si x=tg θ→d ( x )=d (tg (θ ))=sec2θ .dθ

Reemplazando: ∫ csecθ . sec 2θdθ=∫ csecθ (1+tg2θ)dθ

¿∫csecθ .dθ+∫ secθ tgθ dθ

¿∫csecθdθ+∫d (secθ )=ln( tanθ2 )+secθ+C

∫ √ x2+1x

dx=ln( x

√x2+1+1 )+√ x2+1+C

39. Hallar: ∫ 1

x2√4−x2dx

Solución:

1

∫ 1

x2√4−x2dx=1

2∫1

x2√1−( x2 )2d x

senθ= x2→dx=d (2 senθ )=2cosθ .dθ

Reemplazando: ∫ 1

x2√4−x2=1

2∫secθ .2cosθ .dθ

4 sen2θ

¿∫ 1

4 sen2θdθ

¿ 14∫ csc2θd .θ

¿ 14

(−ctgθ )+C

∴∫ √ x2+1x

dx=−14

√1−( x2 )2

x2

+C

40. Hallar: ∫ 1

sen4 xdx

Solución:

Esta integral es de forma ya conocida, así que tenemos:

→∫ 1sen4 x

dx= 123∫

(1+ tg2 x2 )

3

tg4 x2

d (tg x2 )

¿18∫( tg6( x2 )

tg4 x2

+3tg4( x2 )tg4 x

2

+3tg2( x2 )tg4 x

2

+1

tg4 x2

)d (tg x2 )

¿18∫( tg2( x2 )+tg−4( x2 )+3tg−2( x2 )+3)d (tg

x2)

¿ 18 [∫ tg2( x2 )d (tg x

2)+∫ tg−4( x2 )d (tg x

2)+3∫ tg−2( x2 )d ( tg x

2)+3∫ d( tg x

2)]

∴∫ 1

sen4 xdx=

18 [ tg3( x2 )

3−

1

3tg3( x2 )−

3

tg( x2 )+3 tg( x2 )]+C

41. Hallar: ∫ 1

cos4 xdx

Solución:


→∫ 1cos4 x

dx= 123∫

(1+tg2( x2 +π4 ))

3

tg4( x2 + π4 )

d (tg( x2 + π4 ))

∴∫ 1cos4 x

dx=18 ( tg3( x2+ π

4 )3

− 1

3 tg3( x2+ π4 )

− 3

tg( x2 + π4 )

+3 tg( x2 + π4 ))+C

42. Hallar: ∫ 1

sen2 x .cos4 xdx

Solución:


→∫ (1+tg2 x)3−1

tg2 xd (tgx )=∫( tg

4 x

tg2 x+

2 tg2 x

tg2 x+

1

tg2 x )d (tgx )

∫ tg2 xd ( tgx )+2∫ d (tgx )+∫ tg−2 x d ( tgx )

∴∫ 1sen2 x .cos4 x

dx=tg3 x3

+2 tgx− 1tgx

+C

43. Hallar: ∫ 1

cos3 x sen2 xdx

Solución:

∫ 1

cos3 x sen2 xdx=∫ sec3 x . csec2 x .dx¿∫ sec3 x (1+ctg2 x )dx

¿∫ sec3 x .dx+∫ secx . csec2 x .dx¿∫ sec3 x .dx+∫ sec x . (1+ctg2 x ) . dx

¿∫ sec3 x .dx+∫ secx .dx+∫ csecx . ctgx .dx

∴∫ 1

cos3 x sen2 xdx=−cscx+ 3

2ln ( tgx+secx )+ 1

2tgx . secx+C

44. Hallar: ∫1

sen2 x2

cos3 x2

dx

Solución:

∫ 1

sen2 x2

cos3 x2

dx=2∫ csec2 x sec 3 xdx

Por el problema anterior (43):

∴∫ 1

sen2 x2

cos3 x2

dx=2[−cscx+ 32

ln ( tgx+secx )+ 12tgx. secx ]+C

45. Hallar: ∫ 1

sen5 x

Solución:

∫ 1sen5 x

=¿ 116

∫(1+ tg2( x2 ))

4

tg5( x2 )d( tg x

2)¿

¿ 116

∫( tg8 x2

tg5( x2 )+

4 tg6 x2

tg5( x2 )+

6 tg4 x2

tg5( x2 )+

4 tg2 x2

tg5( x2 )+ 1

tg5( x2 ))d (tg x2)

¿ 116 (∫ tg3 x

2d (tg x

2 )+4∫ tgx2d (tg x

2 )+6∫d (tg x

2 )tg

x2

+4∫d ( tg x2 )tg3 x

2

+∫( tg x

2 )tg5( x2 ))

∫ 1

sen5 x=

116 ( tg

4 x2

4+2 tg

2 x2+6 ln|tg x

2|−21

tg2 x2

−1

4 tg4 x2

)+C46. Hallar: ∫ sen5 x 3√cosx dx

Solución:

→∫ sen5 x 3√cosx dx=∫−(1−cos2 x )2cos

13 x (−senx ) . dx=∫ (−cos4 x+2 cos2 x−1 ) cos

13 x d (cosx )=−∫cos

133 x d (cosx )+2∫cos

73 d (cosx )−∫cos

13d (cosx )

∴∫ sen5 x 3√cosx dx=−316

cos163 x+ 2

10.3 .cos

103 x−3

4cos

43 x+C

47. Hallar: ∫ 1( x+a ) ( x+b )

dx

Solución:

Aplicando fracciones parciales tendremos:

∫ 1( x+a ) ( x+b )

dx= 1(b−a )∫

1( x+a )

d (x+a)+ 1a−b∫

1( x+b )

d (x+b)

¿ 1(b−a )

ln ( x+a )+ 1(a−b )

ln ( x+b )∴∫ 1( x+a ) ( x+b )

dx= 1a−b

ln| x+bx+a|+C48. Hallar: ∫ (x2−5 x+9 )

x2−5 x+6dx

Solución:

∫ (x2−5 x+9 )x2−5 x+6

dx=∫ ( x2−5x+6 )x2−5x+6

dx+∫ 3(x2−5x+6 )

dx

¿∫dx+∫ 3

(x2−5x+6 )dx

¿∫dx+3(∫ dxx−3

−∫ dxx−2 )¿∫dx+3(∫ d ( x−3 )

x−3−∫ d (x−2 )

x−2 )

∴∫ (x2−5 x+9 )x2−5 x+6

dx=x+3 ln|x−3x−2|+C

49. Hallar: ∫ 1(x−1)(x+2)(x+3)

dx

Solución:

Por fracciones parciales:

∫ 1(x−1)(x+2)(x+3)

dx= 112∫

d (x−1)(x−1)

−13∫

d ( x+2 )( x+2 )

+ 14∫

d ( x+3 )( x+3 )

∴∫ 1(x−1)(x+2)(x+3)

dx= 112

ln ( x−1 )−13

ln (x+2 )+ 14

ln ( x+3 )+C

50. Hallar: ∫ 2x2+41x−91¿( x−1 ) ( x−3 )(x−4)

dx ¿

Solución:

Por fracciones parciales tenemos:

→∫ (2x¿¿2+41x−91)( x−1 ) ( x−3 ) ( x−4 )

dx=4∫ d (x−1)(x−1)

−7∫ d ( x+3 )( x+3 )

+5∫ d ( x−4 )( x−4 )

¿

∴ ∫2 x2+41x−91¿( x−1 ) ( x−3 )(x−4)

dx ¿=4 ln ( x−1 )−7 ln ( x+3 )+5 ln ( x−4 )+C

51. Hallar: ∫ ( x−7 )(x−1)(x−5) ² (x+2) ²

Solución:

x ²−8 x+7(x−5) ²(x+2) ²

= A(x−5)

+ B

(x−5)2+ C(x+2)

+ D

(x+2)2

x ²−8 x+7=A (x−5)(x+2) ²+B (x+2) ²+C (x−5) ² ( x+2 )+D (x−5)²

Si x=5→B=−849

Si x=−2→D=2749

x2−8 x+7= (A+C ) x3+ (B−A−8C+D ) x2+ (−16 A+4 B+5C−10 D ) x+(4 B−20 A+50C+25 D)

{ A+C=0B−A−8C+D=1

−16 A+4 B+5C−10D=−84 B−20 A+50C+25D=7

→ A= 30343C=−30

343

→∫ ( x−7 )(x−1)(x−5)² (x+2) ²

dx=∫ [ 30343

(x−5)+

−849

(x−5)2 +

−30343

(x+2)+

2749

(x+2)2 ]dx∴∫ ( x−7 )(x−1)

(x−5) ²( x+2) ²dx=

+30343

ln ( x−5 )− 30343

ln ( x+2 )+ 849 (x−5 )

−27

49(x−2)

52. Hallar: ∫ x4dxx4−1

Solución:

∫ x4dxx4−1

=∫( x4−1x4−1

+ 1x4−1 )dx¿∫(1+

1

x4−1 )dx=∫dx+∫ 1

(x4−1 )dx

Por fracciones parciales:

∫ 1

x4−1dx=1

4∫ 1

(x−1 )d ( x−1 )+(−1

4 )∫ 1(x+1 )

d ( x+1 )+(−12 )∫ 1

(x2+1 )dx

→∫ x4dxx4−1

=∫ dx+ 14∫ d ( x−1 )

(x−1 )−1

4∫ d ( x+1 )

( x+1 )−1

2∫ 1

(x2+1 )dx

∴∫ x 4

x4−1=x+ 1

4ln ¿

53. Hallar ∫ 1

x3+1dx

Solución:

Por fracciones parciales tenemos:

∫ 1

(x3+1 )dx=1

3∫1

(x+1)d ( x+1)+ 1

3∫(−x+2 )x2−x+1

dx

¿13∫

1(x+1)

d ( x+1 )−16∫

d( x2−x+1)(x2−x+1)

+12∫

d (x−12)

(x−12 )

2

+(√32 ) ²

13

ln|x+1|−16

ln (x2−x+1 )+ 1

√3arctg

2(x−12 )

√3+C

54. Hallar ∫ 1

x4+1dx

Solución:

1

x4+1= Ax+Bx2+√2x+1

+ Cx+Dx2−√2 x+1

1= (Ax+B ) (x2−√2 x+1 )+(Cx+D)(x2+√2x+1)

1=A x3−A√2 x2+AX+B x2−√2Bx+B+C x3+C√2x2+Cx+D x2+D√2 x+D

1= (A+C ) x3+ (B+D−A√2+C√2 )x2+( A−B √2+C+D√2 ) x+(B+D)

{ A+C=0B+D−√2 A+√2C=0A−√2B+C+√2 D=0

B+D=1

→A=1

2√2; B=

12;C=

−12√2

; D=12

∫ 1x4+1

dx= 12√2 (∫ x+√2

x2−√2 x+1dx−∫ x−√2

x2−√2 x+1dx )

¿ 12√2 (∫ 2x+√2

x2+√2x+1dx−∫ x

x2+√2x+1dx−∫ 2x−√2

x2−√2 x+1dx+∫ x

x2−√2 x+1dx )

¿ 12√2 [∫ 2 x+√2

x2+√2 x+1dx−∫ 2 x−√2

x2−√2x+1dx−1

2 (∫ 2 x+√2x2+√2 x+1

dx−∫ √2x2+√2 x+1

dx )+ 12 (∫ 2 x−√2

x2−√2 x+1dx+∫ √2

x2−√2 x+1dx )]

¿ 12√2 [∫ 2 x+√2

x2+√2 x+1

dx−∫ 2 x−√2

x2−√2x+1

dx−12 (∫ 2 x+√2

x2+√2 x+1

dx−√2∫d (x+ √2

2 )(x+ √2

2 )2

+ 12

)+ 12 (∫ 2 x−√2

x2−√2x+1

dx+√2∫d( x−√2

2 )(x−√2

2 )2

+ 12

)]∴∫ 1

x4+1dx=¿ 1

4√2ln( x2+√2x+1

x2−√2 x+1 )+ 14 [arctg (x+ √2

2 )−arctg( x−√22 )]+C ¿

55. Hallar: ∫ 1

x (x7+1 )dx

Solución:

∫ 1

x (x7+1 )dx=∫( x7+1

x (x7+1 )− x7

x (x7+1 ) )dx¿∫ 1

xdx−∫ x6

(x7+1 )dx

¿∫ 1xdx−1

7∫ d ( x7+1 )

(x7+1 )

∴∫ 1

x (x7+1)dx=lnx−1

7ln (x7+1 )+C

56. Hallar: ∫ 1

x (x5+1 )dx

Solución:

∫ 1

x (x5+1 )dx=∫ [ (x5+1 )

x (x5+1 )− x5

x (x5+1 ) ]dx¿∫ 1xdx−∫ x4

(x5+1 )dx¿∫ 1

xdx−1

5∫ d (x5+1 )

x5+1

∴∫ 1

x (x5+1 )dx=lnx−1

5ln (x5+1 )+C

57. Hallar: ∫ 1

x4 (x3+1 )2dx

Solución:

∫ 1

x4 (x3+1 )2dx=∫ x3+1

x4 (x3+1 )2 dx−∫ x3

x4 ( x3+1 )2 dx

¿∫ 1

x 4 (x3+1 )dx−∫ 1

x (x3+1 )2dx

¿∫ x3+1

x 4 (x3+1 )dx−∫ x3

x4 (x3+1 )dx−[∫ x3+1

x (x3+1 )2dx−∫ x3

x (x3+1 )2dx ]

¿∫ 1x 4 dx−2∫ dx

x (x3+1 )+∫ x2dx

(x3+1 )2

¿∫ 1x 4 dx+

13∫ d (x3+1 )

(x3+1 )2 −2 [∫ (x3+1 )x (x3+1 )

dx−∫ x3

x (x3+1 )dx ]

¿∫ 1x 4 dx+

13∫ d (x3+1 )

(x3+1 )2 −2∫ 1xdx+ 2

3∫ d (x3+1 )

(x3+1 )

∴∫ 1

x4 (x3+1 )2dx= −1

3 x3− 1

3 (x3+1 )−2 ln (x )+ 2

3ln|x3+1|+C

58. Hallar: ∫ x2

( x−1 )10 dx

Solución:

∫ x2

( x−1 )10 dx=∫(x2−1 )( x−1 )10 dx+∫ dx

( x−1 )10¿∫ ( x+1 )

(x−1 )9dx+∫ d ( x−1 )

( x−1 )10

¿∫ dx

(x−1 )8+2∫ d ( x−1 )

( x−1 )9+∫ d ( x−1 )

( x−1 )10 ∴∫ x2

( x−1 )10 dx=−1

7 ( x−1 )7− 1

4 ( x−1 )8− 1

9 ( x−1 )9+C

59. Hallar: ∫ 1

x8+x6dx

Solución:

∫ 1

x8+x6dx=∫ 1

x6 (x2+1 )dx¿∫ (x2+1 )

x6 (x2+1 )dx−∫ x2

x6 (x2+1 )dx¿∫ dx

x6−∫ dx

x4 (x2+1 )

¿∫ dxx6 −[∫ (x2+1 )

x4 (x2+1 )dx−∫ x2

x4 (x2+1 )dx]¿∫ dx

x6−∫ dx

x4+∫ dx

x2 (x2+1 )

¿∫ dx

x6−∫ dx

x4+∫ dx

x2−∫ dx

x2+1∴∫ 1

x8+ x6dx=−1

5x5+ 1

3 x3−1x−arctg (x )

60. Hallar: ∫ 1

(x+1)2 (x2+1 )2dx

Solución:

→Q ( x )=(x+1)2 (x2+1 )2;Q' (x )=2 (x2+1 ) ( x+1 ) (3 x2+2 x+1 )

{Q1 ( x )=mcd (Q (x ) ,Q' (x))=(x2+1 ) (x+1 )→X (x )=A x2+Bx+C

Q2(x )=Q (x )Q 1 ( x )

=(x2+1 ) ( x+1 )→Y (x )=D x2+Ex+F

→∫ dx

( x+1 )2 (x2+1 )2= A x2+Bx+C

(x2+1 ) ( x+1 )+∫ D x2+Ex+F

(x2+1 ) ( x+1 )dx

1

(x+1)2 (x2+1 )2=

(2 Ax+B ) (x2+1 ) ( x+1 )−(3 x2+2 x+1) ( A x2+Bx+C )(x+1)2 (x2+1 )2

+ Dx2+Ex+F(x2+1 ) ( x+1 )

1= (2 Ax+B ) (x3+ x2+x+1 )− (3x2+2 x+1 ) ( A x2+Bx+C )+(D x2+Ex+F ) (x2+1 ) ( x+1 )

1=D x5+(−A+D+E ) x4+(D+E+F−2B ) x3+ (A−B−3C+D+E+F ) x2+(2 A−2C+E+F ) x+(B−C+F)

{D=0

−A+D+E=0D+E+F−2 B=0

A−B−3C+D+E+F=02 A−2C+E+F=0

B−C+F=1

→A=−14

; B=14;C=0 ; D=0 ; E=−1

4; F=3

4

∫ dx

(x+1)2 (x2+1 )2=

−x2

4+ x

4+0

(x2+1 ) ( x+1 )+∫

0− x4+ 3

4(x2+1 ) ( x+1 )

dx¿ −1

4.

x2−x(x2+1 ) ( x+1 )

− 14∫ x−3

(x2+1 ) (x+1 )dx

¿ −14

x2−x( x2+1 ) ( x+1 )

−14 [∫ −2

( x+1 )dx+∫ 2 x

x2+1dx−∫ dx

x2+1 ]∫ 1

(x+1)2 (x2+1 )2dx=

−(x2−x)4 (x2+1 ) (x+1 )

+12

ln ¿

61. Hallar: ∫ dx

(x4−1 )2

Solución:

→Q ( x )=( x4−1 )2 ;Q' ( x )=8 x3 (x4−1 )

{Q1 ( x )=mcd (Q ( x ) ,Q' ( x ) )=(x4−1 )→X ( x )=A x3+B x2+Cx+D

Q2 ( x )= Q ( x )Q1 ( x )

=(x4−1 )→Y ( x )=Ex3+F x2+Gx+H

→∫ dx

(x4−1 )2= A x3+Bx2+Cx+D

(x4−1 )+∫ Ex3+F x2+Gx+H

(x4−1 )dx

1

(x4−1 )2=(3 A x2+2 Bx+C ) (x4−1 )−( 4 x3 ) (x3+B x2+Cx+D )

(x4−1 )2+ Ex3+F x2+Gx+H

( x4−1 )

1=E x7+(F−A ) x6+(G−2B ) x5+(H−3C ) x 4−(4 D+E ) x3−(3 A+F ) x2−(2B+G ) x−(C+H )

{E=0

F−A=0G−2B=0H−3C=04 D+E=03 A+F=02 B+G=0C+H=−1

→A=0 ; B=0;C=−14

; D=0 ; E=0; F=0 ;G=0 ; H=−34

∫ dx

(x4−1 )2=−1

4x

(x4−1 )2+(−3

4 )∫ dx

x4−1

¿−14

x

(x4−1 )2+(−3

4 )[−14 ∫ 1

(x+1)dx+ 1

4∫1

( x−1)dx+(−1

2 )∫ dx

x2+1 ]∴∫ dx

(x4−1 )2=−1

4x

(x4−1 )2+ 3

16ln ( x+1 )−¿ 3

16ln|x−1|+ 3

8arctgx+C ¿

62. Hallar: ∫ dx

(x2+1 )3

Solución:

→Q ( x )=( x2+1 )3 ;Q' ( x )=3 (x2+1 )2 (2x )=6 x (x2+1 )2

{Q1 ( x )=mcd (Q ( x ) ,Q' ( x ) )=(x2+1 )2→X ( x )=A x3+B x2+Cx+D

Q2 ( x )= Q ( x )Q1 ( x )

=(x2+1 )→Y ( x )=Ex+F

→∫ dx

(x2+1 )3= A x3+B x2+Cx+D

(x2+1 )2+∫ Ex+F

x2+1dx

1

(x2+1 )3=(3 A x2+2Bx+C ) (x2+1 )2−4 x (x2+1 ) (A x3+B x2+Cx+D )

(x2+1 )4+Ex+Fx2+1

1=E x5+(F−A ) x4+(2E-2B ) x3+(3 A−3C+2F ) x2+(2 B−4 D+E ) x+(C+F )

{E=¿F−A=0

2E-2B=03 A−3C+2F=02 B−4 D+E=0

C+F=0

→∫ dx

(x2+1 )3=

38x3+ 5

8x

(x2+1 )2 +∫38

x2+1dx

∴∫ dx

(x2+1 )3=1

8( 3x3+5 x )

(x2+1 )2+ 3

8arctgx+C

63. Hallar: ∫ x4−2 x2+2

(x2−2 x+2 )2dx

Solución:

∫ x4−2 x2+2

(x2−2 x+2 )2dx=∫ [1−( 4 x−2

(x2−2 x+2 )2 )+( 4 x−2x2−2 x+2 )]dx

¿∫d x−∫ 4 x−2

(x2−2x+2 )2dx+∫ 4 x−2

x2−2 x+2dx

¿∫dx−∫ 4 x−2

(x2−2 x+2 )2dx+2[∫ 2x−2

x2−2 x+2dx+∫ 1

( x−1 )2+1dx ]

¿∫dx−2 [−∫ d ( 1x2−2 x+2 )+∫ d (x−1 )

[ ( x−1 )2+1 ]2 ]+2∫ d (x2−2x+2 )x2−2 x+2

+2∫ d ( x−1 )( x−1 )2+1

¿∫dx+2∫ d ( 1x2−2x+2 )−2∫ d ( x−1 )

[ ( x−1 )2+1 ]2+2∫ d ( x2−2x+2 )

x2−2 x+2+2∫ d (x−1 )

( x−1 )2+1

∴∫ x4−2 x2+2

(x2−2 x+2 )2dx=x+arctg (x−1 )−

( x−3 )x2−2x+2

+2 ln ( x2−2x+2 )+C

64. Hallar: ∫ x2

√ x2−x+1dx

Solución:

Esta integral tendrá la forma:

∫ x2

√ x2−x+1dx=( Ax+B ) √x2−x+1+λ∫ dx

√ x2−x+1

→x2=A (x2−x+1 )+ 12

(2 x−1 ) ( Ax+B )+λx2=2 A x2+(B−3 A2 ) x+(A−B

2+ λ)

{2 A=1

B−3 A2

=0

A− B2

+ λ=0

→A=12;b=3

4;C=−5

8

→∫ x2

√ x2−x+1dx=( x2 + 3

4 )√x2−x+1−58∫ dx

√ x2−x+1

¿( x2+ 34 )√ x2−x+1−5

8∫ dx

√(x−12 )

2

+12

∴∫ x2

√ x2−x+1dx=( x2+ 3

4 )√ x2−x+1−58

ln(x−12+√ x2−x+1)+C

65. Hallar: ∫ x5

√1−x2dx

Solución:

Esta integral tendrá la forma:

∫ x5

√1−x2dx=( A x4+B x3+C x2+Dx+E )√1−x2+ λ∫ dx

√1−x2

x5

√1−x2x5=(4 A x3+3B x2+2Cx+D )√1−x2−

x ( A x4+B x3+C x2+Dx+E )√1−x2

+ λ

√1−x2

x5=(−5 A ) x5+ (−4 B ) x4+(4 A−3C ) x3+(3 B−2D ) x2+(2C−E ) x+ ( λ+D )

{−5 A=1−4 B=0

4 A−3C=03 B−2 D=02C−E=0λ+D=0

A=−15

;B=0 ;C=−415

; D=0 ; E=−815

; λ=0

∴∫ x5

√1−x2dx=(−1

5x4− 4

15x2− 8

15 )√1−x2

66. Hallar: ∫ x6

√1+ x2dx

Solución:

La integral tendrá la forma:

∫ x6

√1+ x2dx=( Ax5+B x4+Cx3+Dx2+Ex+F )√1+x2+ λ∫ dx

√1+x2

x6=6 A x6+5B x5+(5 A+4C ) x4+( 4B+3D ) x3+(3C+2 E ) x2+ (2D+F ) x+(E+ λ )

{6 A=15B=0

5 A+4C=04 B+3 D=03C+2 E=02D+F=0E+λ=0

A=16; B=0 ;C=−5

24; D=0 ; E= 5

16; F=0 ; λ=−5

16

→∫ x6

√1+x2dx=( x5

6−5x3

24+ 5 x

16 )√1+x2− 516

∫ dx

√1+x2

∴∫ x6

√1+ x2dx=( x5

6−5 x3

24+ 5 x

16 )√1+x2− 516

ln (x+√1+ x2)+C

67. Hallar: ∫ dx

x5√ x2−1

Solución:

Diremos que x=1t→d ( x )=−1

t 2d (t)

→∫ dx

x5 √x2−1=−∫ t 4dt

√1−t 2

La integral ∫ t 4dt

√1−t 2tendrala forma :

→∫ t 4dt

√1−t 2=(At 3+Bt 2+Ct+D )√1−t2+λ∫ dt

√1−t 2

t 4=( 3 A t 2+2Bt+C ) (1−t 2 )−t (A t3+B t2+Ct+D )+λ

A=−13

;B=−13

;C=−12

;D=29; λ=1

2

→∫ dx

x5 √x2−1=−∫ t 4dt

√1−t 2=−[(−t 3

3− t 2

3− t

2+ 2

9 )√1−t2+ 12∫ dt

√1−t2 ]∴∫ dx

x5√ x2−1=−[(−1

3 x3 −1

3 x2 −1

2 t+ 2

9 )√1− 1x2 +

12arcsen( 1

x )]+C68. Hallar: ∫ dx

( x+1 )3√ x2+2x

Soluciones:

Diremos que x+1=1t→d ( x )=−1

t2d (t)

→∫ dx

( x+1 )3 √x2+2x=−∫ t2dt

√1−t 2¿∫√1−t 2dt−∫ d t

√1−t 2

∴∫ dx

( x+1 )3√ x2+2x= 1

2 ( x+1 ) √1− 1

( x+1 )2−1

2arcsen( 1

x+1 )+C

69. Hallar: ∫ x2+x+1

x √x2−x+1

Solución:

∫ x2+x+1

x √x2−x+1=∫ xdx

√x2−x+1+∫ dx

√ x2−x+1+∫ dx

x √ x2−x+1

¿12 [∫ 2 x−1

√x2−x+1dx+∫ dx

√ x2−x+1 ]+∫ dx

√x2−x+1+∫ dx

x √x2−x+1

¿ 12∫

2x−1

√x2−x+1dx+ 3

2∫dx

√(x−12 )

2

+(√32 )

2−∫ dt

√(t−12 )

2

+(√32 )

2

Donde : x=1t

∴∫ x2+x+1

x√ x2−x+1=√x2−x+1+ 3

2ln(x−1

2+√x2−x+1)−ln( 1

x−1

2+√( 1

x−1

2 )2

+ 34 )+C

70. Hallar: ∫ x3 (2x2+1 )−32 dx

Solución:

Aplicando el criterio de Chebichev a la integral tendremos:

m=3 ;n=2 ; p=−32

;a=1 ;b=2→m+1n

∈Z ;1+2x2=z2

→∫ x3 (2 x2+1 )−32 dx=1

4∫(1− 1

z2 )dz∴∫ x3 (2x2+1 )−32 dx=1

4 (z+ 1z )+C=1

4 ( 2 x2+2

√1+2 x2 )+C71. Hallar: ∫ x3 (2x2+1 )3dx

Solución:

∫ x3 (2x2+1 )3dx=∫ x3 [8 x6+1+6x2 (2x2+1 ) ]dx¿∫ (8 x9+x3+12x7+6 x5 )dx

¿8∫ x9dx+12∫ x7dx+6∫ x5dx+∫ x3dx

∴∫ x3 (2x2+1 )3dx=8 x10

10+3 x8

2+ x6+ x4

4+C

72. Hallar: ∫ 1

x4 √1+x2dx

Solución:

Diremos que x=1t→dx=−1

t 2dt

→∫ 1

x4 √1+ x2dx=−∫ t 3

√1+t 2dt¿−[ ( At2+B t+C )√1+x2+λ∫ dt

√1+t 2 ]t 3=(2 At+B ) (t 2+1 )+( A t 3+B t 2+Ct )+λ

Resolviendo obtendremos:

∴∫ 1

x4 √1+x2dx=−[( t 2

3−2

3 )√t 2+1]+C=−[( 13 x2 −

23 )√ 1

x2 +1]+C73. Hallar: ∫ dx

x .3√ x5+1

Solución:

∫ dx

x .3√ x5+1

=∫ x−1 (x5+1 )−12

→m=−1 ;n=5 ; p=−13

; a=1 ;b=1

Por el criteriodeChebichev :m+1n

∈Z→ x5+1=z¿= 3 z2dz

55√( z3+1 )4

→∫ dx

x .3√x5+1

=35∫

z . dz

( z−1 ) ( z2+z+1 )

z

( z−1 ) ( z2+z+1 )= Az+B

z2+z+1+ Cz−1

z=( A+C ) z2+(B+C−A ) z+(C−B )

{ A+C=0B+C−A=1C−B=0

A=−13

;B=13;C=1

3

35∫

z . dz

( z−1 ) ( z2+z+1 )=

35∫

−13

z+ 13

z2+z+1dz+

35∫

13

z−1dz

¿− 110 (∫ 2 z+1

z2+z+1dz−3∫ 1

(z+12 )

2

+(√32 )

2 dz )+ 15∫ 1

z−1dz

∴∫ dx

x .3√ x5+1

=−110

ln ( z2+z+1 )+ 25√3

arctg( 2 z+1

√3 )+ 15

ln (Z−1 )+CSiendo z= 3√x5+1

74. Hallar: ∫1

x2 (x3+2 )53

dx

Solución:

Dando la forma adecuada para aplicar el criterio de Chebichev:

∫ x−2 (x3+2 )−53 dx→m=−2 ;n=3 ; p=−5

3;a=2 ;b=1

Si :m+1n

+ p=−2∈Z→2 x−3+1=z3dx=−[ 213 . z2

( z3−1 )13 ( z−1 ) ]dz

Reemplazando:→∫ 1

x2 (x3+2 )53

dx=−14∫ z3−1

z3 dz¿−14∫ dz+ 1

4∫ z−3dz

∴∫ 1

x2 (x3+2 )53

dx=−14

. 3√ 2x3 +1−1

8.

1

3√ 2

x3 +12 +C

75. Hallar: ∫ 1

√ x3 .3√1+

4√x3dx

Solución:

Dando la forma adecuada para aplicar el criterio de Chebichev:

∫ x−3

2 (1+x34 )

−13dx→m=−3

2; n=3

4; p=−1

3;a=1 ;b=1

Si :m+1n

+ p=−1∈Z→1+x−3

4 =z3dx=−[ 213 . z2

( z3−1 )13 ( z−1 ) ]dz

Reemplazando:→∫ 1

√ x3 .3√1+

4√x3dx=∫−4 z . dz

∴∫ 1

√ x3 .3√1+

4√x3dx=−2 z2+C=−2 3√ 1

4√ x3+1+C

76. Hallar: ∫ 13+5cosx

dx

Solución:

Diremos que tg( x2 )=t ; senx= 2 t

t 2+1;dx= 2

t2+1dt

→∫ 13+5cosx

dx=∫ 2dt

(t 2+1 ) [3+5( 1−t 2

1+t 2 )]¿∫1

22−t2dt

∴∫ 13+5cosx

dx=14

ln|2+t2−t|+C= 1

4ln|2+ tg( x2 )

2−tg( x2 )|+C77. Hallar: ∫ 1

senx+cosxdx

Solución:


t 2+1;dx= 2

t2+1dt

→∫ 1senx+cosx

dx=∫ 2

( t2+1 )( 2 tt 2+1

+ 1−t 2

1+t 2 )dt

¿−2∫ 1

t 2−2 t−1dt¿−2∫ 1

(t−1 )2− 2√22dt

∴∫ 1senx+cosx

dx=−1√2

ln|t−1−√2t−1+√2 |+C= 1

√2ln|tg( x2 )−1+√2

tg( x2 )+1−√2|+C78. Hallar: ∫ cosx

1+cosxdx

Solución:


t 2+1;dx= 2

t2+1dt

→∫ cosx1+cosx

dx=2∫ ( 1−t 2)

( 1+ t2 )(1+ 1−t 2

1+t 2 ) (1+t 2)dt

¿−∫( t 2−1t 2+1 )dt¿−∫ dt+2∫ 1

t2+1dt

∴∫ cosx1+cosx

dx=−t+2arctg ( t )+C=−tg( x2 )+x+C

79. Hallar: ∫ senx1−senx

dx

Solución:

∫ senx1−senx

dx=−∫ senxsenx−1

dx¿−[∫ senx−1senx−1

dx+∫ 1senx−1

dx ]¿−[∫dx+∫ 1

senx−1dx ]


t 2+1;dx= 2

t2+1dt

→−[∫ dx+∫ 1senx−1

dx ]=−[∫ dx+∫ 2

(t 2+1 )( 2 tt2+1

−1)dt ]¿−[∫dx−2∫ 1

(t−1 )2dt ]

∫ senx1−senx

dx=−x− 2t−1

+C=−x− 2

tg( x2 )−1+C

80. Hallar: ∫ 18−4 senx+7cosx

dx

Solución:


t 2+1;dx= 2

t2+1dt

→∫ 18−4 senx+7cosx

dx=2∫ 1(t−5 ) ( t−3 )

dt

Por fracciones parciales obtendremos:

2∫ 1( t−5 ) (t−3 )

dt=∫ dtt−5

−∫ dtt−3

∴∫ 18−4 senx+7cosx

dx=ln|t−5|−ln|t−3|+C=ln|tg( x2 )−5

tg( x2 )−3|+C81. Hallar: ∫( 1+tg ( x )

1−tg (x ) )dxSolución:

∫( 1+tg ( x )1−tg (x ) )dx=−[∫( tg (x )−1

tg (x )−1 )dx+2∫ 1tg ( x )−1

dx ]¿−[∫dx+2∫ cosxsenx−cosx

dx ]Diremos que tg( x2 )=t ; senx= 2 t

t 2+1;dx= 2

t2+1dt

−[∫ dx+2∫ cosxsenx−cosx

dx]=−[∫ dx+2∫ 2 (1−t 2 )

(t2+1 )2( 2 tt 2+1

−1−t 2

1+t 2 )dt ]

¿−[∫dx−4∫ (t2−1 )(t2+1 ) (t 2+2t−1 )

dt ]Por fracciones parciales obtendremos:

−[∫ dx−4∫ (t 2−1 )(t 2+1 ) ( t2+2 t−1 )

dt ]=−{∫ dx−4 [−14∫ 1

t+1+√2dt+ 1

2 ( 12∫ 1

t 2+1d (t 2 )+∫ 1

t 2+1dt)−1

4∫ 1

t+1−√2dt ]}

∴∫( 1+ tg ( x )1−tg ( x ) )dx=−x−ln|( t+1 )2−2|+ ln|t 2+1|+2arctg (t )+C¿ ln| tg( x2 )

2

+1

( tg( x2 )+1)2

−2|+C82. Hallar: ∫ 1

1+3 cos2 xdx

Solución:

Diremos que tg (x )=t ; senx= t

√t 2+1;dx= 1

t 2+1dt

∫ 1

1+3cos2 xdx=∫ dt

(t 2+1 )(1+3

t2+1 )¿∫1

t2+4dt

∴∫ 1

1+3 cos2 xdx=1

2arctg ( t2 )+C=1

2arctg ( tg ( x )

2 )+C

83. Hallar: ∫ 1

3 sen2 x+5 cos2 xdx

Solución:

∫ 1

3 sen2 x+5 cos2 xdx=∫ 1

3+2 cos2 xdx

Diremos que tg (x )=t ; senx= t

√t 2+1;dx= 1

t 2+1dt

→∫ 1

3+2cos2 xdx=∫ 1

(t 2+1 )(3+2

t 2+1 )dt

¿∫ 1

3t 2+5dt

∴∫ 13 sen2 x+5 cos2 x

dx= 1√15

arctg (√3√5

t)+C¿ 1√15

arctg (√3√5

tg ( x ))+C

84. Hallar: ∫ 1

sen2 x+3 senx .cosx−cos2 xdx

Solución:

∫ 1

sen2 x+3 senx .cosx−cos2 xdx=∫ 1

3 sen (2x )−2cos (2x )d (2 x )

¿∫ 13 sen (M )−2cos (M )

d (M )

Dónde: M=2 x

Diremos que tg( M2 )=t ; senM= 2 t

t 2+1;dM= 2

t 2+1dt

→∫ 13 sen (M )−2cos (M )

d (M )=∫ 2

(t2+1 )( 6 tt 2+1

−2 (1−t 2 )

1+t 2 )dt

¿∫ 1

2t 2+6 t−2dt

¿ 12∫

1

(t+ 32 )

2

−(√134 )

2dt

∴∫ 1sen2 x+3 senx. cosx−cos2 x

dx=−√1313

ln|2. tgx+3+√132. tgx+3−√13|+C

85. Hallar: ∫ 1(2−senx ) (3−senx )

dx

Solución:


t 2+1;dx= 2

t2+1dt

→∫ 1(2−senx ) (3−senx )

dx=∫ 2

(2− 2 t

t 2+1 )(3−2 t

t 2+1 )( t2+1 )dt

¿∫ 1

(t2−t+1 ) (3 t 2−2 t+3 )dt

¿−∫ t−1

t 2−t+1dt+∫ 3 t−2

3 t 2−2 t+3dt

∴∫ 1(2−senx ) (3−senx )

dx=−√2

2arctg( 3√2tg( x2 )−√2

4 )+ 2√33

arctg (2√3 tg( x2 )−√3

3 )+C86. Hallar: ∫ 1−senx+cosx

1+senx−cosxdx

Solución:

∫ 1−senx+cosx1+senx−cosx

dx=−∫ senx−cosx−1senx−cosx+1

dx¿−[∫dx−2∫ 1senx−cosx+1

dx]


t 2+1;dx= 2

t2+1dt

→−[∫ dx−2∫ 1senx−cosx+1

dx ]=−[∫dx−2∫ 1

t2+tdt ]

¿−[∫dx−2∫ 1

(t+12 )

2

−( 14 )

2 dt ]∴∫ 1−senx+cosx

1+senx−cosxdx=−[ x−2 ln( t

t+1 )]=−[x−2. ln( tg( x2 )tg( x2 )+1 )]+C

Integrales indefinidas

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