Integrales indefinidas
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Transcript of 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:
Esta integral es de forma ya conocida, así que tenemos:
→∫ 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:
Esta integral es de forma ya conocida, así que tenemos:
→∫ (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:
Diremos que tg( x2 )=t ; senx= 2 t
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:
Diremos que tg( x2 )=t ; senx= 2 t
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 ]
Diremos que tg( x2 )=t ; senx= 2 t
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:
Diremos que tg( x2 )=t ; senx= 2 t
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:
Diremos que tg( x2 )=t ; senx= 2 t
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]
Diremos que tg( x2 )=t ; senx= 2 t
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