Transcript
Cálculo Diferencial e Integral - Dos métodos de integración.
Prof. Farith J. Briceño N.
Objetivos a cubrir
Código : MAT-CDI.8
Integración : Integrales por sustitución trigonométrica. Integración : Integrales por descomposición en fracciones simples. Ejercicios resueltos
Ejemplo 1 : Integre
Z
p
dx x2 5
Solución : Hacem os el cambio trigonom étrico x=
p
5 sec t;
dx =
p
5 sec t tan t dt
la integral se transform a en Z
p
dx x2
5
=
Z
p
r
Z Z p Z p Z p 5 sec t tan t dt 5 sec t tan t dt 5 sec t tan t dt sec t dt = ln jsec t + tan tj + C; = = = p p p 2 2 5 sec t 5 5 tan t 5 (sec t 1)
5 sec t tan t dt = p 2 5 sec t 5
com o x=
p
5 sec t
x sec t = p 5
=)
=)
cos t =
Para calcular tan t en función x nos p odem os ayudar con el triángulo rectangular, con a =
x = a sec t
=)
Por Pitágoras
p
x2
tan t = Z
p
dx x2
x = ln p + 5 5
p
x2 p
5
a2
así,
Luego
5 c:a: = x hip
x a
sec t =
c:o: =
p
p
5 5
c:o: = c:a:
+ C1 = ln
p
x+
donde, C = C1
ln
x2 p
5 5
:
p
x2 p 5
p
5
+ C1 = ln x +
p
x2
5 + C;
5: F
Ejemplo 2 : Integre
Z
p
x2 dx
4
Solución : Hacem os el cambio trigonom étrico x = 2 sen t;
dx = 2 cos t dt
la integral se transform a en Z
p
4
x2 dx =
Z q
4
(2 sen t)2 2 cos t dt = 2
=2 donde,
así,
com o
Z
2
cos t dt = Z p
x = 2 sen t
4
Z
Z
p
4 (1
Z
p
4
4 sen2 t cos t dt
sen2 t) cos t dt = 4
Z p
cos2 t cos t dt = 4
Z
cos2 t dt;
1 + cos 2t t 1 t 1 dt = + sen 2t + C1 = + sen t cos t + C1 ; 2 2 4 2 2
x2 dx = 4
=)
t 1 + sen t cos t + C1 2 2 sen t =
x c:o: = 2 hip
1
= 2t + 2 sen t cos t + C;
=)
t = arcsen
x 2
Para calcular cos t en función x nos p odem os ayudar con el triángulo rectangular, con a = 2
x = a sen t
=)
Por Pitágoras
sen t = p
c:a: =
a2
x a
x2
así, cos t = es decir,
Z p
Luego
Z p
Z
Ejemplo 3 : Integre
x 2
x2 dx = 2 arcsen
4
+2
x 2
c:o: = hip p
p
2 x 2
x2 dx = 2 arcsen
4
2
x2
4
x2
4
;
+ C = 2 arcsen
x 2
+
p x 4 x2 + C: 2
p x 4 x2 + C: 2
+
F
dx p x x2 + 3
Solución : Hacem os el cambio trigonom étrico x=
p
3 tan t;
dx =
p
2
3 sec t dt
la integral se transform a en Z
dx = p x x2 + 3
Z
p
3 sec2 t dt r p p 3 tan t 3 tan t
= 2
+3
Z
sec2 t dt = p tan t 3 tan2 t + 3
Z
sec2 t dt = p tan t 3 sec2 t
Z
sec2 t dt 1 = p p tan t 3 sec t 3
Z
sec t dt; tan t
donde, Z
sec t dt = tan t
1 Z Z 1 cos t dt = dt = csc t dt = ln jcsc t sen t sen t cos t
Z
así,
Z
com o
p
1 dx = p ln jcsc t p x x2 + 3 3
cot tj + C;
cot tj + C;
x c:o: tan t = p = c:a: 3 p Para calcular csc t en función x nos p odem os ayudar con el triángulo rectangular, con a = 3 x=
x = a tan t
=)
Por Pitágoras
3 tan t
tan t = p
hip =
hip = c:o:
Luego
Z
x a
x2 + a2
así, csc t =
=)
p
3 + x2 x
y
dx 1 = p ln p x x2 + 3 3
p
cot t =
3 + x2 x
p 3 1 : = x tan t
p
3 + C: x F
Ejemplo 4 : Integre
Z
2x2 (x
6x + 7 2
1) (x + 2)
dx
Solución : Observem os que el grado del p olinom io del num erador, 2, es m enor que el grado del p olinom io del denom inador, 3, p or lo tanto, no se dividen los p olinom ios. Adem ás, el denom inador ya está factorizado.
2
Escribim os las fracciones sim ples corresp ondientes 2x2
6x + 7
A
=
1)2 (x + 2)
(x
x
1
+
B 1)2
(x
C : x+2
+
Buscam os los valores de las constantes A, B y C, 2x2 (x
6x + 7
A
=
2
1) (x + 2)
x
B
+
1
+
2
(x
1)
C A (x = x+2
1) (x + 2) + B (x + 2) + C (x
1)2
2
(x
1) (x + 2)
;
de aquí, 2
2x
6x + 7 = A (x
2
1) (x + 2) + B (x + 2) + C (x
1) :
Para obtener los valores de las constantes le dam os valores arbitrarios a x Si x = 1, entonces
2 (1)
Si x =
2
6 (1) + 7 = A ((1)
1) ((1) + 2) + B ((1) + 2) + C ((1)
1)
2
=)
3 = 3B
B=1
=)
2, entonces
2 ( 2)
2
6 ( 2) + 7 = A (( 2)
1) (( 2) + 2) + B (( 2) + 2) + C (( 2)
1)
2
=)
27 = 9C
=)
C=3
Si x = 0, entonces 2 (0)
2
6 (0) + 7 = A ((0)
com o B = 1 y C = 3, se tiene que 7 =
1) ((0) + 2) + B ((0) + 2) + C ((0)
Entonces
2x2
6x + 7 1)2 (x + 2)
(x p or lo tanto,
Z
A=
2A + 2 (1) + (3), de aquí,
2x2
6x + 7 2
(x
1) (x + 2)
dx =
Z
A
=
x
1
1 x
+
dx +
1
1)
=)
7=
2A + 2B + C;
1:
B 1)2
(x Z
2
+
1 (x
1)
2
C ; x+2
dx +
Z
3 dx: x+2
La prim era integral del lado derecho de la igualdad se resuelve haciendo el cambio de variable u=x así,
Z
1 x
1
1;
Z
dx =
du = dx;
du = u
ln juj + C1 =
ln jx
1j + C1 :
La segunda integral del lado derecho de la igualdad se resuelve haciendo el m ism o cambio de variable u=x y la integral se transform a en
Z
1 (x
1)
2
1;
dx =
Z
du = dx;
du = u2
1 + C2 = u
1 x
1
+ C2 :
La tercera y últim a integral del lado derecho de la igualdad la resolvem os haciendo el cambio de variable u = x + 2; se obtiene
Z
Finalm ente
Z
3 dx = 3 x+2
2x2 (x
6x + 7 1)2 (x + 2)
Z
du = dx;
du = 3 ln juj + C3 = 3 ln jx + 2j + C3 : u
dx =
ln jx
1j
1 x
1
+ 3 ln jx + 2j + C: F
Ejemplo 5 : Integre
Z
x2 + 8x + 14 dx (2x + 4) (x2 + 2x + 2)
Solución : Observem os que el grado del p olinom io del num erador, 2, es m enor que el grado del p olinom io del denom inador, 3, p or lo tanto, no se dividen los p olinom ios. Adem ás, el denom inador ya está factorizado, puesto que, el p olinom io p (x) = x2 + 2x + 2 no es factorizable en los núm eros reales.
3
Escribim os las fracciones sim ples corresp ondientes x2 + 8x + 14 A Bx + C = + 2 : (2x + 4) (x2 + 2x + 2) 2x + 4 x + 2x + 2 Buscam os los valores de las constantes A, B y C, A x2 + 2x + 2 + (Bx + C) (2x + 4) x2 + 8x + 14 A Bx + C = + 2 = ; (2x + 4) (x2 + 2x + 2) 2x + 4 x + 2x + 2 (2x + 4) (x2 + 2x + 2) de aquí, 2
2
x + 8x + 14 = A x + 2x + 2 + (Bx + C) (2x + 4) : Para obtener los valores de las constantes le dam os valores arbitrarios a x Si x =
2, entonces
2
2
( 2) + 8 ( 2) + 14 = A ( 2) + 2 ( 2) + 2 + (B ( 2) + C) (2 ( 2) + 4)
=)
2 = 2A
A=1
=)
Si x = 0, entonces 2
2
(0) + 8 (0) + 14 = A (0) + 2 (0) + 2 + (B (0) + C) (2 (0) + 4)
=)
14 = 2A + 4C;
C=3
com o A = 1, se tiene que 14 = 2 (1) + 4C, de aquí,
Si x = 1, entonces 2
2
(1) + 8 (1) + 14 = A (1) + 2 (1) + 2 + (B (1) + C) (2 (1) + 4)
p or lo tanto,
23 = 5A + 6B + 6C;
B = 0:
com o A = 1 y C = 3, se tiene que 23 = 5 (1) + 6B + 6 (3), de aquí,
Entonces
=)
x2 + 8x + 14 1 3 = + 2 ; (2x + 4) (x2 + 2x + 2) 2x + 4 x + 2x + 2 Z
x2 + 8x + 14 dx = (2x + 4) (x2 + 2x + 2)
Z
1 dx + 2x + 4
Z
3 1 dx = x2 + 2x + 2 2
Z
dx +3 x+2
Z
dx : x2 + 2x + 2
La prim era integral del lado derecho de la igualdad se resuelve haciendo el cambio de variable u = x + 2; así,
Z
dx = x+2
Z
du = dx;
du = ln juj + C1 = ln jx + 2j + C1 : u
Para la segunda integral del lado derecho de la igualdad com pletam os cuadrado 2
2
x + 2x + 2 = (x + 1) + 1
Z
=)
dx = x2 + 2x + 2
Z
dx (x + 1)2 + 1
;
hacem os el cambio de variable u = x + 1; y la integral se transform a en
Luego
Z Z
dx = x2 + 2x + 2
Z
du = dx;
du = arctan u + C2 = arctan (x + 1) + C2 : u2 + 1
x2 + 8x + 14 1 dx = ln jx + 2j + 3 arctan (x + 1) + C: (2x + 4) (x2 + 2x + 2) 2 F
Ejercicios 1. Calcular las siguientes integrales haciendo la sustitución trigonométrica apropiada. Z Z Z Z dx dt dx d p p p 2: 3: 1: 4: 7 + 2t2 16 x2 4 3x2 9+ 4
2
5:
Z
p
dx 3x2 2
6: 11:
15: 19: 23: 27: 31:
35:
39: 43: 48: 52: 56:
60: 64: 68: 72:
76: 80: 84:
88:
Z
Z
Z
Z
Z
Z
Z
Z Z
Z
Z Z
Z
Z
Z Z
Z Z
Z
Z
Z
dx ax2
Z
y dy
Z
y 2 dy
Z
dx
Z
p
1 + t2 dt a (y 2 + 4) (y 2 + 4) Z Z Z p cos x dx dx dx 12: 13: 14: 2t t2 dt 2 2 3=2 sen x 6 sen x + 12 (1 + x2 ) (4x2 25) Z Z Z p p x2 dx p p et 9 e2t dt 16: 5 4t t2 dt 17: dx 18: 2 2 5 x x + 4x + 5 Z Z p 2 Z 3x 9x 4 dx dx p p p 20: dx 21: dx 22: 4 2 2 x x x 2 x + 2x + 5 4x x2 Z Z Z x dx t dt ex dx dx p p p p 26: 25: 24: 4x x2 a4 t4 16 + 6x x2 1 + ex + e2x Z Z Z p 2x + 1 2x 1 sen t cos t 2t dx 28: dx 29: e 9 dt 30: dt x2 + 2x + 2 x2 6x + 18 9 + cos4 t Z Z Z sec2 2x ln x dx 3x2 x3 p p dx 32: 33: dx 34: dx 2 2 2x + 5 9 + tan 2x 7 + x2 x 1 4 ln x ln2 x p p Z Z Z x tan7 x + tan5 x dx e3x dx p 36: dx 37: 38: 2 dx 4 4 5=2 2 2 tan ( =4) + tan x x x 5 (x + 6) (e x + ex ) Z Z Z e 3x dx x2 dx dx dx p p p 40: 41: 42: 3=2 3 6 2 8x x 13 9 x 3 x2 (e2x 9) Z Z Z Z dx dx dx dx dx p p p 46: 47: 44: 45: 16 + x2 2 + x2 x2 4 x2 5 x x2 + 3 Z Z p Z dx dx 1 x2 dx p p p 49: 50: dx 51: 2 2 2 2 x 9x + 6x 8 x 1 x x + 2x + 5 p Z Z Z x2 dx x2 a2 dx p p dx 53: 54: dx 55: 4 3=2 2 3 x2 2 2 x x x + 9 x 16 (a x ) Z Z Z p p p dx 57: 1 4r2 dr 58: t 4 t2 dt 59: x2 4 9x2 dx 5=2 2 (5 4x x ) Z Z Z p 2x 1 t3 dt x dx p p p dx 61: 62: 63: x2 9 x2 dx x2 4x + 5 t2 + 4 1 x2 Z Z Z e2x dx dx dx dx p p p p 65: 66: 67: 2x 4x 2 2 2 2 2 1+e +e x 16x 9 x x +9 x + 4x + 8 Z Z Z 2 4x dx sen x dx x dx p p p 69: 71: dx 70: 2 2 2 2 7 + 5x cos x + 4 cos x + 1 x + 6x x + 6x p Z Z Z dx dx sen 2x sen x x p dx 74: 75: dx 73: 2 2 3=2 2 2 sen x + 5 x 8x + 19 (x 4) x (1 x ) Z Z Z x 1 x dx e2x dx 5=2 p p 77: dx 78: 79: x2 1 dx x x 3=2 + 4 2 6e 6e x x x 8x + 3 Z Z Z dx dx dx x dx p p 83: 81: 82: 3x2 x + 1 x4 4x2 + 3 x+x+2 x x2 Z Z Z p dx dx dx x3 4 9x2 dx 85: 86: 87: 2 x2 + 2x x2 + 2x + 5 (x2 + 2x + 2) Z Z Z p sec2 x dx x2 dx dx p p 91: 89: t t2 dt 90: x2 6x + 10 2 + 3x 2x2 tan2 x 2
p
b
7:
5=2
8:
5=2
5
9:
p
bx2
10:
5t
92:
96:
100:
104:
108:
112: 116: 119:
Z
Z
p
x2
+ 2x + 5 dx dt p
93:
Z
Z
p
t2
+ 1 dt
94:
x2 dx p 9 x2
Z
Z p
t2 4 dt t2
Z
Z
dx
p
x2
+ px + q
2y + 1 p dy (t + 1) + 2t y2 + 9 Z Z Z p ax dx dx p 1 2t t2 dt 101: 102: 103: 3=2 2 1 + a2x 1 6x x2 (y + 4) Z Z Z Z 3x 6 x2 dx y dy (2x + 1) dx p p p dx 105: 106: 107: 2 2 4 x2 + 2x + 2 x 4x + 5 4x x 16 9y Z Z Z Z 2x 1 t2 dt 3x dx 2x 3 p p dx 109: 110: dx 111: 2 2 2 2 x2 6x + 18 4 x x + 2x + 5 (t + 1) Z Z Z Z 2x 1 dz y 2 dy (2x 8) dx p p p 115: dx 113: 114: 5=2 2 1 x x2 x2 4x + 5 z 1 z2 (9 y ) Z Z Z p dt sen 2x + cos x p 2 x x2 dx 117: dx 118: 2 x + sen x 2 sen 2 t 2t + 26 Z Z Z dt dt 2t dt p p p 120: 121: 2 2 2 16 + 6t t 16 + 4t 2t t 2t + 26 t2
97:
98:
sen x dx 16 + cos2 x Z y 3 dy
95:
99:
2. Calcular las siguientes integrales utilizando descomposición en fracciones simples. Z 2 Z Z Z Z x +1 dx 2 dx 3t2 6t + 2 5t + 3 1: dx 2: 3: 4: dt 5: dt x2 x x2 x 2 x2 + 2x 2t3 3t2 + t t2 9 Z Z Z 4 Z Z 3x3 dx 2 dx x + 8x2 + 8 x2 dx dx 6: 7: 8: dx 9: 10: 3 2 x2 + x 2 x2 1 x3 4x (x + 1) x2 (x 1) Z Z Z Z 5x2 + 6x + 9 dx dt x3 4x x3 + x 11: dx 12: 13: 14: 2 2 2 2 2 dx (x 3) (x 3) (x + 1) t2 (t + 1) (x2 1) Z Z Z Z x3 + 1 dx dx x2 + 19x + 10 dx 18: 15: 16: 17: 2 2 2 dx 4 2 4 2 9x + x x + x2 + 1 (x 4x + 5) (x 3) (2x + 1) Z Z Z Z x4 + 1 dx x2 2x 1 x3 dx x2 dx 19: 20: dx 21: 22: 2 2 x4 + x2 (x2 + 3) (x2 + 1) (1 + x2 ) (x3 + 4x) Z Z 4 Z Z 4x2 + 3x + 6 x dx dx x+1 24: dx 25: 23: 26: dx 2 2 2 x4 1 x3 1 (x4 + x2 + 1) (x2 + 2) (x2 + 3) Z Z Z Z dx x+4 dx (x 6) dx 29: 27: dx 28: 30: 2 3 + 3x`2 2 x (x2 + 4) x x2 2x (x + 1) (x + x + 1) Z 3 Z Z Z t2 + 2 dt x +x+1 dx 9 dx 31: dx 32: 33: 34: 2 3 x (x + 1) (x 1) (x + 2) (x + 3) 8x + 1 t (t2 + 1) Z Z Z Z 2x3 x x2 3 dx dx 35: dx 36: dx 37: 38: x4 x2 + 1 x3 + 4x2 + 5x + 2 x3 1 x3 + x2 + x Z Z Z Z 5x3 + 2 dx (x + 2) dx (20x 11) dx dt 42: 39: 40: 41: 3 3 2 2 2 x 5x + 4x (3x + 2) (x 4x + 5) x2 (x2 1) (t + 1) Z Z Z Z x3 8x2 1 dx (x 11) dx dx 5x 2 43: 44: 45: 46: dx x2 + 3x 4 (x + 3) (x 2) (x2 + 1) 2x3 + x x2 4 Z Z Z Z x2 dx dt dx 18 dx 48: 47: 49: 50: 2 2 3 2 4 2x + 9x + 12x + 4 16x 1 (t + 2) (t + 1) (4x2 + 9)
6
51: 55:
59: 63: 67: 71: 75:
Z
Z Z
Z
Z
Z
Z
Z
(17x 3) dx 3x2 + x 2
52:
(t + 3) dt 4t4 + 4t3 + t2
56:
60:
dt (t + a) (t + b)
64:
4x dx 16
Z
68:
cos x dx sen x + sen3 x
72:
(5x + 7) dx x2 + 4x + 4
Z
Z
Z
76:
x4 + 1 dx
Z
61:
2x3 + 5x2 + 16x dx x5 + 8x3 + 16x
65:
x (3 Z
ln x)
x2 + 3x + 3 dx 3 x + x2 + x + 1
Z
Z
62:
4
4 + 5x2 dx x3 + 4x Z e5x dx
58:
x6 dx x2 16
Z
Z
77:
ex dx e4x 1
x2
Z
54:
dx
Z
73:
3x2 + 7x dx 3 x + 6x2 + 11x + 6
Z
Z
69:
dx x3 + 1
Z
57:
2x2 3x 36 dx (2x 1) (x2 + 9)
dx ln x) (1
Z
53:
2x2 + 41x 91 dx (x 1) (x + 3) (x 4)
Z
(4x 2) dx x3 x2 2x
5x3 x4
x2 4x 4 dx x3 2x2 + 4x 8
66:
t3 dt t3 8
70:
t2 dt 4 t 8t
74:
Z
2
(e2x + 1)
t2 + 2 dt t (t2 1) Z
x2 dx x2 + x 6
t3 1 dt 4t3 t
Z
x 3 dx x3 + x2
30x2 + 52x + 17 24x3 dx 9x4 6x3 11x2 + 4x + 4
Z
x4 + 3x3 5x2 4x + 17 dx 2 x3 + x2 5x + 3 x (x2 + 1) Z Z Z 2 2x3 + 9x dx (3x 13) dx x 5x + 9 81: 82: 83: dx x2 + 3x 10 (x2 + 3) (x2 2x + 3) x2 5x + 6 Z Z Z 2 x2 8x + 7 5x2 11x + 5 x +x+2 84: 85: dx 86: dx 2 dx 3 2 + 5x 2 x 4x 2 x2 1 (x 3x 10) Z 4 Z Z 2x2 + x 8 2x 3 x 6x3 12x2 + 6 dx 89: dx 87: dx 88: 2 3 2 x 6x + 12x 8 x3 + 4x (x2 3x + 2) Z Z Z 2 6x2 + 22x 23 x + 2x 1 5x2 + 3x 2 90: dx 91: dx 92: dx 3 2 2 x + 2x (2x 1) (x + x 6) 27x3 1 Z Z Z 2 sec2 x + 1 sec2 x 2x2 x + 2 x 4x + 3 93: dx 94: dx 95: 2 dx x5 2x3 + x 1 + tan3 x x (x + 1) Z Z Z 6x2 2x 1 2x2 + 3x + 2 (x 2) dx 96: dx 97: dx 98: 3 3 2 4x x x + 4x + 6x + 4 2x2 + 7x + 3 Z Z Z dx 3x + 5 (2x + 21) dx 100: 99: 101: 2 dx 2 2 + 4x + 5) 2 (x 4x + 3) (x 2x2 + 9x 5 (x + 2x + 2) Z 2 Z Z x + 19x + 10 3x2 21x + 32 3x2 x + 1 102: dx 103: dx 104: dx 4 3 3 2 2x + 5x x 8x + 16x x3 x2 Z Z Z 2x4 2x + 1 2x2 + 13x + 18 2t2 + t 4 105: dx 106: dx 107: dt 5 4 3 2 2x x x + 6x + 9x t3 t2 2t Z Z Z x2 + x dx 5x2 3x + 18 x3 108: 109: dx 110: dx 3 2 3 4 x x +x 1 9x x x + 2x2 Z Z Z x2 + 3 x dx 2x2 x + 2 111: dx 112: 113: dx 3 2 3 2 x +x 2x x + 2x + x + 2 x5 + 2x3 + x 78:
79:
x2 3x 7 dx (2x + 3) (x + 1)
80:
Respuestas: Ejercicios 1:1: 1:5: 1:9:
arcsen p
3 3
ln
arcsen
1 4x
p
+ C;
3x + p pb x a
p
1:2:
arcsen
3x2
2 + C;
+ C;
1:10:
p
3 2 x
+ C;
1:6: 5 3
p
1 + t2
b b
1:3: ln
3=2
p
ax +
+ C;
p
14 14
p
arctan
p
14 7 t
ax2
b + C;
1:11:
1 25
7
p
+ C;
1:4:
1:7: x 4x2
25
+ C;
1 24
ln
+
p
y2 + 4 1:12:
p
2
+ 9 + C;
3=2
3 3
+ C;
arctan
1:8: senpx 3 3
1 12
y3
(y2 +4)3=2
+ C;
+ C;
1:13:
1 2
arctan x +
1:16:
9 2
arcsen
1:19:
1 4
1:22:
p
x2 x
t+2 3
2
x
1:29: 3 e2t
1:41:
arcsen
1:46: 1:50:
p
2 2
p
1 243
p
ln
1:59:
2 27
1:61:
p 1
p
2 2 x
1:76:
1 2
ln x2 p
p
x2
x 4
p
x4
x2
4x
3=2
+
2
1 2
1:90:
ln tan x +
p
2
4 135 x
1:92: 2 ln x + 1 + p
arcsen arcsen
p 1:104: 3 x2
t2
p x 2
t+1 p 2
t+1 p 2
p
1
1+
arcsen
1:85:
1 2
(x + 1)
+ C;
p
p
2 4
p
5 5
1:95:
p
t2
t
1 3
1 4
1:98: t2 + C;
2t
2+
p
1 4
arctan 1:101:
x
1 2
1:108:
p 2 1
4x + 5 + C; x2
x + C;
1:118: p
1:121: 2 t2
1 4
1:83:
+ C;
4x
ln
1:89:
1:116:
1 z
arcsen
5 3
q
1+
8
t2
+ C;
14 + C;
arcsen
p
3 3 1 4
1:45:
x3 + C;
arctan 1 x
1:49: x a
4
p
1
x 4
+ C;
x2 + C;
+ C;
t2 t2
4
x2
x3 8
4 + C;
4x + 5 + C;
p
x2 + C;
9
p 2 35 25
4 5x
p
35 7 x
arctan
p
p
5 5
2 5 arctan
+ C;
sen x + C;
8x + 19 + C; 1 x 6e
+ C;
p
p 2 x+1
7 7
arctan
x+1 2
arctan
arcsen (2t
1 2
+ C;
+ C;
2t 1 4
1) +
p
ln t +
2 +8
p
t2 + C;
t
t2 + 1 + C; 1 t+1
arccos
p 1
+ C;
1:106: x
arctan
+ C;
1 3
1 6
1:103:
arcsen
3 2 4y
arcsen
+ C;
p
10 10
(x + 3) + C;
+ C; 1 2
1:109:
arctan t
t 2t2 +2
+ C;
x2 + 2x + 5 + C;
z2 z2
2x+1 3
2 + C; p
3 3
+ C; 1 2
y 2 +4
3 ln x + 1 +
9 8
p
ex +1 ex 1
1:96:
1:102: py
6x + 18 +
ln sen2 x + sen x 2t + 26 + 2 ln t
1 8
arctan
6x + 10 + C;
x2 (6 + x) + C;
ln
p
1 3
+
x2
p 2 7 7
1:86:
+ C; x
x2 + 9 + C;
p
3 1
1 3
p p 1:99: 2 y 2 + 9 + ln y + y 2 + 9 + C;
+ C;
1:113:
ln
x2 x2
p 1:93: 12 t t2 + 1 +
ln x2
1 12
x+x+2
p x2 + px + q + C;
1:111: 3 2x + x2 + 5 x2
p
ln
cos x + C;
p
1:78:
3) + 3 ln x2
arctan ax ln a 1 2
x + C;
+
5 3
+ C;
+ C;
1:69:
4 3
t2 a2
arcsen
x2 + 7 x2
5 + C;
1:72: 2 sen x x
p
arcsen
2+
x p 1
9x
ln
ln jx + 2j + C;
4x 3 5
arcsen
p
x2 + C;
2t + 26 + C; + C;
2 2
1 2
9
1 2
8 + C;
x2 16 x2
2 3=2
1:66:
1:75:
1:80:
ln jxj
x 8
x2
x2
4x + 5 + 3 ln x x 3
p
1:58:
x2 + 6x + C;
1) + C;
ln 2x + p + 2
1:105: 6 arcsen
(2x + 1)
p
1 2
p
p
4 arctan
(6x
p
1 32
+
1 3
1:40:
x a2
2 3x
2 arccos
5 + C;
9x2 + 6x
4r 2 + C;
1
arcsen
+ C;
2x + x2 + 5 + C;
1) + ln x2 + 2x + 2 + C; x 2
11 11
p
cos2 x + 4 cos x + 1 + C;
1 + C;
p
1:88:
x2
1 x
4 x
x2
ln x +
4
+ C;
+ C;
1:52: p
9 + C;
p
ln x + 3 +
1:91: x + 8 arctan (x
x2 + C;
4x + 5 + 3 ln x
9 arcsen
2 2
4
4x + 5 + C;
3 arcsen
ln t
t2 t
9
+
arctan ( x
p 1:112: 2 x2
p
+ C;
2 + C;
x2
arctan
+ C;
x2 + 2x + 5 +
4
x 3
32 1215
x+1 4x+2x2 +4
ln x +
p
81 8
16x2
p
1
p
e2x 9 e2x
e2t + C;
9
6x + 18 +
1:34:
1 5x
1:44:
arccos
1:60: 2 x2
1 9x
p 1:77: 2 x
p 2 11 11
1:82:
tan2 x
p
5 16
+ ln
x2
1 2r
q
9x2
1:25:
ln x
tan 2x 3
p
p
2
p
x2 + 4x + 5 + x + 2 + C;
+ C;
+ C;
ln 3x + 1 +
1 128
1:55:
ln cos x + 2 +
1 1
8x2 + 3 + C;
1) + C;
arctan (x + 1) +
ln t +
1:74: p
1 3
p
p
2 729
3 5
arctan
10 5 x
x2 + 2x + 5 + C;
1:63:
1:71:
1 6
1:37:
e2x e2x 9
x
p
ln
1 t 2e
+
1:21:
1:28:
p
4 + C;
arcsen (2r) +
x2 + C; 1:65:
arctan
1:48:
9x2 + C;
4
x2 + 6x + C;
+ C;
p
q
x2
+ C;
1 4
1:57:
1:68:
26x + 33
1 5x
1:87:
p
3=2
p
+ C;
ln x + 1 +
+ C; p x3
1:18:
arcsen
1:31:
p 3 10 4
1 729
ln x +
3 x
x2 a2 x2
1 3a2
1:24:
et 3
arcsen
x2 + 2x + 5 + x + 1 + C;
+ C;
3
e2x 9 e2x
p
x2 +3 x
1:51:
x+2 5
2
4
9x2
4
p x
1 8x
1:84:
1:120:
+ C;
9x2
4
ln
x2 + 4x + 8 + C;
4
p
p
p
p
x2 + C;
5
arctan tan2 x + C;
1 2
1:43:
p
4+
arcsen (2x
1:117:
3 3
1:54:
3 ln x + 3 +
1:81:
1:115:
1 81
x 72
p x 2 p x+2
ln
1:110:
p
1 2x
+ C;
3 2x
1:33:
q
1 2187
8 + C; 1:62: 1 p + 1 + 2 e4x + e2x + 1 + C;
2x
x2 + 6x
1 4
1:107:
1 x2 x
p
3 ln
cos2 t 3
arctan
tan2 x
+ C;
t2 + 4 t
1:73:
1:100:
3 3 x
2
1:70:
9 2
+
p
arcsen
+ C;
3x 2
p
1:97:
3 x
4x
ln x + 2 +
1:94:
1 x
1 2
1:39:
1:47:
p
1 6
4 ln x + C;
+ C;
+ C;
(x+2)3 x2 )3=2
(5
1:67:
1:79:
1 e2x +1
5 5 x
9 2
1:15:
arctan (x + 1) + ln x + 2x + 2 + C;
1:30:
1:36:
p
p
arcsen
t2 + C;
2t
2
1:27:
+ C;
p
1)
ln 1 + 2ex + 2 1 + ex + e2x
ln2 x
1
1:42:
9+x2 x
ln 2e
x 48
p
+ C;
arcsen
3
1 2
x 3
x2 + ln
1:56:
x x+6
6
1
1
1 3
p
(e2x +1)3
arctan
1:53:
1:64:
px
arctan
1 q 3
p
2+ln p x 5
5 2
1:17:
(t
p
x2 + C;
4x
1 2
1) +
p 1:20: 3 x2 + 2x + 5
+ C; 1:23:
p
arcsen (t
t2 + C;
4t
3 arccos et + C;
9
1:38:
5
+ C; 2
2 arcsen 6 6
p
1 2
1:14:
3=2
2 x3
2
p
1:35:
2 2
1:26: 2 arcsen
p
x2
+ C;
t+2 2
+
1 2 x
arcsen
1:32:
x 1 2 x2 +1
+
+ C; 1 4
(2x + 1)
1:119:
p
arcsen
2t + 26 + C;
1 27
1:114: 2 3 5
3
y 9
y2
+ C;
x2 + C;
x
t
p
+ C;
2:1: x
ln jxj + 2 ln jx
1j + C;
x 2 x+1
2:2:
1 3
2:6:
3 2 2x
2:9:
ln jx + 1j +
2:12: 2:15:
ln
15 2
arctan (x 1 x
+ C;
1 x
x+2 1 3 x2 +x+1
2 ln jx
2:33:
3 2
ln j2x + 1j
2:35:
1 2
ln x4
ln jxj
2:54:
x x2 +4
ln
ln x
2:70:
1 4t
2:72:
5 2
+
3 2
3 2
ln
3 x
+ C;
x2
2:90: 2 ln jxj + 5 162
1 x
p
+
+ C;
2:88:
2 81
ln 9x + 3x + 1
2:97: 2 ln jx + 2j 3x+1 x2 1 3x3
p 2 3 3
3 3 x
arctan
1 2 2x
3j
ln jx
ln j3x 3 3
1j +
1j +
1j + C;
arctan x + ln jx
1j + C;
p 5 3 27
(2 tan x
7 130
16 3 3 x
3 1
3 x+2
p
2 2
p
3 3
2:106: 2 ln jxj
2:109: 2 ln jxj
1 x2 +1
ln jxj +
(x
ln jx
2:95: 1 2
1 x
1
1 3
p
3 3
(2x + 1) + C;
4j + C;
2 x
ln jxj +
3 2
(x 1)3 x+1
ln
+ C;
ln 2x2 + 1 + C;
ln j2x + 1j + C;
5 3
ln j3x
2j + C;
1) + C; 7 ln jx + 3j + 5 ln jx
1j
4j + C;
2 ln jx + 1j + C;
2x
1
2:63: 4 5
b
ln jx
9 5
2j p 2 3 3
2
ln t + 2t + 4
t+a t+b
ln
a
+ C;
ln jx + 3j + C; p
3 3
arctan
(t + 1) + C;
p
3 6
p
3 3
arctan
(t + 1) + C;
1 2x
3 ln jx + 1j +
1 4
ln j2x + 3j + C;
x x
3 2
+ C;
2:86: x + 2 ln jx 2 ln jxj +
1 2
ln jx + 1j + C;
1j
1 2x
arctan
+ C;
1j + C;
x3 (x+1)2
1 4
2:107: 2 ln jtj
+
8 x+1
3 4
+ C;
2:99:
1j
2:96:
ln jxj
arctan (x + 1) +
1 x
2:104:
ln jt + 1j + ln jt 2:110:
5 4
ln jx
2:101: 2 ln j2x
+ C;
4 ln jx + 3j + C;
9
1 18
ln x2
2:79:
+ C;
2
3j
1 2
2:66: x +
2:93: 2 ln jxj
x
+
(2x
2:89: 2 ln x + 4
4j +
ln jx + 3j + C;
arctan
ln jx
2:42:
ln jxj
2
ln
161 6
+ C;
2j
+ C;
ln x + 4x + 5 + C;
3 ln jx
3 3
3 3
1j +
4 3x+6
p
1 4
2 ln jx + 1j + 3 ln jx + 3j + C;
ln j2x + 1j + C;
+ C;
ln jx
2:83: x + 3 ln
2j
+ C;
+ C;
+ C;
p
ln x2 + x + 1
+ C;
ln jt
x 1 x+1
ln
2j + C;
1) + C;
2:103: 2 ln jxj + ln jx 1 x+3
2 3
1 4
(2x + 1) + C;
ln t + 1 + C;
ln t2 + 2t + 4 +
(6x + 1) + C;
ln jx + 3j
arctan (x + 2) +
x 4 x+4
1 3x
2 ln jtj + C;
1
3 3
+ C;
ln sen2 x + 1 + C;
ln jx + 3j + ln j2x
2j
1 65
ln t
arctan x +
(2x + 1) + C;
1 2x2 +2
2
2:59:
2
3 2
1 12
1j + 3 ln jx
1) + C;
2:98:
+ C;
3 3
p
arctan
ln jx + 2j +
2:56: 4 ln jx
+ C;
+ 512 ln
2j
1 2
1 3
2:45:
arctan
3 9
x+3 x
ln
1 2
1 3t3 +5t 8 (t2 +1)2
2:76: 2 ln jx + 2j
88x 139 (x 2)2
arctan
3 3
p
p
arctan
ln x + 1 +
1) +
1j
7 3
ln jxj
p
1 ex 2 e2x +1
ln jt
2:78:
1 9
ln jx + 2j +
2:69: t +
2:81: 4 ln jx + 5j
2j +
1 2
3 6
+ C;
ln jx
1 6
p
2
1 2
2:51: 4 ln jx + 1j +
1 2
+ C;
arctan
4 9
ln jsen xj 1 6
2:73:
1 2x
1 12
+ C;
+ C;
1) +
2:25: x
1j
+ C;
1 5 5x
+
(2x
(2x
2:29:
arctan t +
2:62:
+ C;
2:71:
2:91: 3 ln jx
ln jxj + ln j2x + 5j + C;
+ ln j2x
+
p 7 2 4
24 ln jx
arctan (x + 1) + C; 1 20
ln
ln x ln x
2:85: 2 ln jx
+ C;
p
+ C;
1j + C;
3 + C;
+ 3 ln jx + 2j + C;
2
ln jx
2 ln jx
p
arctan
151 19x 49(x+2)(x 5)
ln
ln x2 + 1 + C;
ln j3x + 2j
3 2
x 2 x+2
ln j2t + 1j + C;
ln x + 2x
1
1 2
2:68:
2
ln jtan x + 1j +
1 52
1 4
arctan (ex ) +
2:65: 256x +
2:75: 5 ln jx + 2j + 2 3
2x + 3
1 3x+2
+ C;
9 16
1j
ln jx + 1j +
3
x 5 x+2
2:61:
ln x + 4 + C;
1 2
x
1 2x
1 4
3 3
+ C;
ln jx
11t+3 2t2 +t
11 ln jtj + 11 ln j2t + 1j 3 2
1 3
30 x 3
3j
2:21:
2:34: 2 ln jtj
2:48: 2 3x
3 3
p
2j + 2 ln jx + 3j + C;
arctan
p
arctan
2:32:
4 + C;
307x+143 196x2 490x 294
arctan
ln x2 + x + 1 +
1 6
ln jx + 1j
1 2
3j + 2 ln jt + 3j + C;
+ C;
arctan
2:39: 5x +
+ C;
1 6
+
3 9
+
3 8
+ 28 ln jx
arctan x + C;
2:37:
2:41:
1 t+2
x 4x2 +9
3 6
(2x + 1) + C;
+ C;
+
p
p
1) + C;
ln jx
2:58: ex
1j + C;
ln j2t
1 3
2:53: 2:55:
arctan
t+1 t+2
2:50:
2
7 16
1 2 2x
30 343
2:108:
+
28x+17 1 3 (3x+2)(x 1)
2:84:
2:105:
2j + C;
2 ln j2x
4 +
x x+1
ln x2
2:102:
+ C;
arctan (ex ) + C;
1 2
5 1 2 x2 +4
2:82:
2:100:
2x 1 2x+1
ln
ln jx
arctan x +
2:80: 2x +
2:94:
1 8
3 3 x
+ 1 + C;
ln j3x + 2j + C;
4x + 5
p
(2x + 1) + C;
ln
+ ln jtj
2:74: 4 ln
arctan
2:47:
ln x + 9
2:67:
3 3
2 x+1
2j + 3 ln jx + 2j + C;
2
2
2 1 2 ln x p 3 3 (4x
arctan
p
3 3
arctan
arctan x
ex 1 ex +1
2:64:
2:92:
ln x2
ln jxj + 2 ln x + 4 + C;
2:60:
2:87:
3 3
2
3 2
2:77:
p
p 3 3 2
p
1 2 2x
2x+1 x 3
ln
ln x2 + 4 +
1 2
ln jxj
2:44:
ln x2 + 4
1 4
p 5 3 9
129 343
3x3 +6x2 +7x+15 2(x2 +2)(x2 +3)
2 2 x
2 ln jxj + 7 ln x2
3 2x2 2
1 +
x2 +x+1 x2 x+1
ln
1 2 2x
x+1 +
arctan
1 4
+
2 arctan 2:27:
ln x2
p
1j + C;
arctan 2x +
2:57:
4
ln jx + 2j +
2:36:
ln x + x + 1
2 ln jx
2:52:
2x + 1 +
2
1 2
+
2:31: x + ln jxj
ln 4x
2:43: 3 ln jx + 4j
1 4
p 15
ln x + x + 1 +
2
3 4
2) +
2:46: 2 ln jx
+
p 2 3 3
ln x2
1 2
2:5: 3 ln jt
1j + C;
2:11: 6x +
2:16:
1 4
ln j2t
2:8:
+ C;
+ C;
1 2
1j +
+ C;
2:14:
x x3 1 6 x4 +x2 +1
2
1 2
2j + C;
2:40: 4 arctan (x
2:49:
3 3 x
x2 + 1 + C; 1 2
ln
2:23:
p
+ C;
x2 +3 x2 +1
ln x2 + x + 1 + C;
1 3
2:28: ln jx + 1j + 2:30: 3 ln jxj
+ C;
3 3 arctan
1j
1 2
2:20:
1 2x x2 x
+
1
ln x2 + x + 1
1 4
2:18:
p
x2 +3 x2 +2
ln jx
1 2x
x x
ln jt
x 1 x+1
ln
3x 17 2x2 8x+10
4x + 5 +
+ C;
arctan
2:7:
2t+1 t2 +t
t+1 t
ln x2
1 2
2) +
1 16
+
2:4: 2 ln jtj
+ C;
2:10: 2 ln
2:13: 2 ln
2 arctan x
2:24: 3 ln
2:38:
+ C;
3 arctan 3x + C;
x 8x2 +32
2 3
x x+2
ln
1j + 8 ln jx + 2j + C;
4x+3 2x2 +4x+2
5x 3 2x 3
2:19: x
2:26:
2:3:
3x + ln jx
x2
2:17:
2:22:
+ C;
1 2
1j
+ 3 ln jx
1 x 4 2 x2 1
ln jx + 1j + 1 4
ln j2x
x x2 +2x+2
1j +
2j + C; ln x2 + 2 + C;
ln j2x + 1j + C;
1 2x2 +4x+4
ln jx + 5j + C; 1j + C;
3 4
+ C;
+ C;
4 3
ln jx
1j
2:113: 2 ln jxj
1 2
2:111:
3 2
ln jxj +
arctan x
7 6
ln jx + 2j + C; 2
ln x + 1
x 2x2 +2
2:112:
1 5
arctan x
2 5
ln jx + 2j +
1 5
ln x2 + 1 + C;
+ C;
Bibliografía 1. Purcell, E. - Varberg, D: “Cálculo con Geometría Analítica". Novena Edición. Prentice Hall. 2. Stewart, J.: “Cálculo". Grupo Editorial Iberoamericano.
Cálculo Diferencial e Integral - Integración u-sustitución.
Prof. Farith Briceño e-mail :
[email protected]
Última actualizacón: Enero 2010
10