2

INTEGRALES INMEDIATAS Ó PSEUDOINMEDIATAS
Resolver las siguientes integrales.
1.

2.

3.

∫x

5

x 5+1
x6
+C =
+C
5 +1
6

dx =

dx

∫ x4

x −4+1
x −3
−1
+C =
+C =
+C
− 4 +1
−3
3x 3

= ∫ x − 4 dx =

1

2
2
∫ x x dx = ∫ x ⋅ x 2 dx = ∫ x

2+

1
2 dx

=∫

4.

5.

6.

∫

3

x 2 dx = ∫ x

dx

∫5x

∫

=∫x

−1

3 dx

5 dx

=

=

2
+1
x3

5

2
+1
3

1
− +1
x 5

1
− +1
5

x
5

+C =

4

x
+C =
4

5
+1
x2

=5
+1
2

3

+C =

8.

9.

2
x 2
+C =
x7 + C =
7
7
2

3

5

+C =

5

33 5
3 3
x + C = x x2 + C
5
5

55 4
x +C
4

3x 2
3
3 x 2+1
x3
dx = ∫ x 2 dx =
+C =
+C
5
5
5 2 +1
5
1

7.

7

+C =

2 3
x x +C
7

=

2

5
x 2 dx

∫

+1

3



1
2
x2
x 2
+C = 5
+C =
5x 3 + C =
5x dx =  5x = {
5 ⋅ x  = 5 x 2 dx = 5
1
3
3


cte
+1
2
2
2
= x 5x + C
3

∫3

∫

dx
8x 2

∫ 3x

5

=

2 dx

1
3

=

8

∫x

−2

5
+13x 2

5
+1
2

2
− +1
x 3

1

1 x 3
x
3 dx =
+C =
+ C = 33 + C
3
3
2
1
8
8 − +1
8
3
3

+C =

1

3x
7

7

2

2

+C =

6 72
x +C
7

10.

4x
∫ (3 − 4x )dx = ∫ xdx − ∫ 4xdx = 3x −

11.

∫ (x

2

2

2

+ C = 3x − 2 x 2 + C

)

− 7 x + 12 dx = ∫ x 2 dx − ∫ 7 xdx + ∫ 12dx =

1

x 3 7x 2
−
+ 12 x + C
3
2

12.

∫ (x

)

x 4 12 x 3 14 x 2
x4
−
+
− 5x + C =
− 4 x 3 + 7 x 2 − 5x + C
4
3
2
4
No hace faltadescomponer la integral en suma de integrales.
3

− 12 x 2 + 14 x − 5 dx =

(

)

13.

 5
4 
x6
4 x −1
x6
4
5
−2


−
+
x
7
dx
=
x
−
7
+
4
x
dx
=
−
7
x
+
+
C
=
− 7x − + C
∫
∫
2 
6
−
1
6
x
x 


14.

∫  6x



2



+ 4x −

=

6 8
+
x x3

(

)


6
1

dx = ∫  6x 2 + 4x + 8x −3 − dx = ∫ 6x 2 + 4x + 8x −3 dx − 6∫ dx =
x
x



6 x 3 4x 2 8x −2
4
+
+
− 6 ln x + C = 2x 3 + 2x 2 −
− 6 ln x +C
3
2
−2
x2

(

)

 x 3 7x 2 4 
dx x 3 7 x 2
x 3 − 7x 2 + 4
−
+ dx = ∫ x 2 − 7 x dx + 4∫
dx = ∫ 
=
−
+ 4 ln x + C
 x
x
3
2
x
x 
x


15.

∫

16.

∫ (x

17.

∫ (x + 4)

2

)

(

2

)

− 5 dx = ∫ x 4 − 10x 2 + 25 dx =
3

x 5 10 x 3
−
+ 25x + C
5
3

dx Integral de tipo potencial compuesta que responde a la primitiva:

∫f

n

( x ) ⋅ f ' ( x )dx =

f n +1 ( x )
+ C ∀n ≠ −1
n +1

Donde f (x) = x+ 4; f’(x) = 1; n = 3

∫ (x + 4)

3

18.

∫

(x − 1) dx =
2

2

x

9

5

dx = ∫ (x + 4 )3 ⋅1dx =

 x4
2x 2
1

−
∫  12 12 + 12
x
x
x

(x + 4)4
4

+C


dx =  x 7 2 − 2x 3 2 + x − 1 2 dx

∫ 




1

x 2 2x 2 x 2
2
4
2
4
=
−
+
+C =
x9 −
x5 + 2 x + C = x4 x − x2 x + 2 x + C =
9
5
1
9
3
9
3
2
2
2
2
=
x x 4 − 6x 2 + 9 + C
9

(

19.

∫ (2x + 3)

3

9

(

)

)

5
3
1
1 
 7
x dx = ∫ 8x 3 +36 x 2 + 18x + 27 ⋅ x 2 dx = ∫  8x 2 + 36 x 2 + 18x 2 + 27 x 2  ⋅ dx =


7

5

3

8x 2 36x 2 18x 2 27 x 2
16 x 9 72 x 7 36 x 5
=
+
+
+
+C =
+
+
+ 18 x 3 + C =
9
7
5
3
9
7
5
2
2
2
2
4
3
2
 8x 4 36 x 3 18x 2

16x x 72 x x 36x x
=
+
+
+ 18x x + C = 2 x 
+
+
+ 9x  + C


9
7
5
7
5
 9


20.

∫

 3 x 2 − 3 x + 12 dx =





5

∫

3

1

 23
x 3 3x 2
 x − 3x 2 + 12 dx =
−
+ 12x +C =
3
5


2
3
33 5
=
x − 2 x 3 + 12x + C
5

2

21.

22.

∫ (1 + x )(1 + x )dx = ∫ (1 + x + x
2

∫

 3x
2
+
dx = 
1
1

x
x 2 x 2

3x + 2

∫

2

)

+ x 3 dx = x +


dx =



x2 x3 x4
+
+
+C
2
3
4
3

∫

1

−1 
 12
3x 2 2 x 2
 3x + 2 x 2 dx =
+
+C =
1
3


2
2

= 2 x 3 + 4 x + C = 2 x x + 4 x + C = 2 x (x + 2) + C

23.

∫

∫(

)

1

5x x 2 + 7 dx = 5 x 2 + 7 2 x dx . Integral deltipo potencial compuesta, responde a la

primitiva:
f n +1 ( x )
+ C ∀n ≠ −1
n +1
f(x) = x2 + 7; n = ½; f ’(x) = 2x.
Para completar la derivada, solo falta una constante, por lo que multiplicamos dentro
por 2 y fuera por ½.
n
∫ f ( x ) ⋅ f ' ( x )dx =

∫ 5x
24.

2

∫(

33

x 4 − 12 dx =

2

x + 7 dx = 5 x + 7

∫x

∫ (x

)

4

1

5
=
2

2 x dx

)

∫ (x

2

+7

)

1

2 2 x dx

(

)

3

5 x2 +7 2
5
=+C =
3
2
3
2

(x

2

+7

)

3

+C

1

− 12 3 x 3 dx . Integral del tipo potencial compuesta.

f n +1 ( x )
+ C ∀n ≠ −1
n +1
Donde: f(x) = x4 − 12; f ’(x) = 4x3; n = 1/3. La completar la derivada y poder aplicar la primitiva
se multiplica dentro por 4 y fuera por 1/4

∫f

∫ (x

1
4

25.

∫ x(2x

2

−7

)

99

4

− 12

)

1

n

( x ) ⋅ f ' ( x )dx =

(

)

4

(

)

4
1 x 4 − 12 3
3 3 4
=
+C =
x −...