Cálculo Diferencial e Integral Dos métodos de integración. Prof

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
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