Integrales Definidas

/Ed'Z >^

∫ x dx
5

5
∫ x dx =

∫ (x +

x 5+1
x6
+k =
+k
5 +1
6

x )dx
1

3

+1

x1+1 x 2
x2 x 2
x 2 2 x3
( x + x )dx = ∫ ( x + x1 / 2 )dx =
+
+k =
+
+k =
+
+k =
∫
3
1+1 1
2
2
3
+1
2
2
=

x 2 2x x
+
+k
2
3

 3 x x
 dx
−
4
x


∫


^

6 x−

12
x x +k
10
1
− +1

3

+1

1

5

1
3
−
3
x x
1
x2
1 x2
x2 1x2

 dx = ∫ (3x 2 − x 2 )dx = 3 ⋅
−
−⋅
+ k = 3⋅
−⋅
+k =
∫ x 4 
1
1 45
4
43


− +1
+1
2
2
2
2

=6 x−

1 x5
1
+ k = 6 x − x2 x + k
25
10

x 2 dx
∫x

22
x x +k
5

^
3
3

5

+1

x 2 dx
x2
x2
x2
2 x5
2x 2 x
= ∫ 1 dx = ∫ x 2 dx =
+k =
+k =
+k =
+k
∫x
3
5
5
5
+1
x2
2
2
1


∫ x


2

1


∫ x


2

+


+ 2  dx
xx


+


+ 2  dx = ∫ ( x −2 + 4 x

xx


4

−

^

4

−

3
−
2

+ 2)dx =

18
−
+ 2x + k
x
x
− 2 +1

3
− +1
2

x
x
+4
+ 2x + k =
3
− 2 +1
− +1
2

1

x −1
x2
1
1
18
=
+4
+ 2x + k = − − 8 ⋅ 1 + 2x + k = − −
+ 2x + k
1
−1
x
x
x
−
x2
2

∫

dx
4
x

44 3
x +k
3

^

1

1
− +1

1

3

3

−
dx
x4
x4
4x 4
4= ∫ x 4 dx =
+k =
+k =
+k =
∫4 x
1
3
3
3
− +1
4
4

2

2 1


∫  x + 3 x  dx



∫

2

2
1
 x + 3  dx =


x


∫

1

1533
1
3
⋅ x + ⋅ x + 3. x 3 + K = ⋅ x 5 + ⋅ 3 x 8 + 3. 3 x + K =
5
4
5
4

=

∫

8

2

1 5 3 23 2
⋅ x + ⋅ x . x + 3. 3 x + K
5
4

ln x
⋅ dx
x

12
L x+k
2

^

ln( x)
1
∫ x ⋅ dx = ∫ ln( x) ⋅ x ⋅ dx ={∫

}

(Ln( x)) 2
f ⋅ f ' dx =
+k
2
α

sen 3 x
+k
3

∫

sen 2 x ⋅ cos x ⋅ dx

∫


sen 2 x ⋅ cos x ⋅ dx == 


∫

cos 3 x ⋅ sen x ⋅ dx

∫

cos 4 x
cos x ⋅ sen x ⋅ dx = − cos x ⋅ ( −sen x ) ⋅ dx =
+k
4
f3
f'

∫

x x 2 + 1 ⋅ dx

∫
∫

^

∫

∫

1
x x + 1 ⋅ dx =
2
xdx
2x 2 + 3

 sen 3 x
f 2 ⋅ f ' dx  =
+k
3

^

3

2

1

1
5
25
3
3
2
−
−
 x + x 3  dx = ( x 4 + 2 x 3 + x 3 ).dx = x + 2 ⋅ x + x + K =


8
1
5


3
3
8

=

x3 + k

x5 3 2 3 2
+ x x + 33 x + k
54

^

∫

4

−

3

1
( x 2 + 1) 3 + k
3

^

∫

cos 4 x
+k
4

3
2

( x 2 + 1) 3
1 ( x + 1)
+k =
+k
2 x ⋅ ( x + 1) dx = ⋅
3
2
3
f'
1/ 2
f
2
2

1
2

2

1
2x 2 + 3 + k
2

^
2

1

∫
∫xdx

1
=
2x 2 + 3 4
xdx
2x 2 + 3

∫
∫
∫

x 2 dx

2
4

∫


f'

=
dx =
2
2 2x + 3  2 f


1
2

∫

4 xdx

x 2 dx

x 2 dx

=

∫

1 ( 2 x 2 + 3) 2
1
4 x ⋅ ( 2 x 2 + 3) dx = ⋅
+k =
2x 2 + 3 + k
1
4
2
f'
f −1 / 2
2
−

2
x3 +1 + k
3

^

x +1
3

1

f + k =
2x 2 + 3 + k
2


1

x3 + 1

x3 + 1

=

∫

=

2
3

−1
1 ( x 3 + 1) 2
2
x 2 ( x 3 + 1) dx =
3 x 2 ( x 3 + 1) 2 dx = ⋅
+k =
x3 + 1 + k
1
3
3
3
2
−

∫

∫

∫
∫

1

∫


f'

=
dx =
2 x3 +1  2 f

3 x 2 dx

x ⋅ ( x 2 + 1) 4 dx

^

∫ cos x ⋅ sen

2
4
∫ x ⋅ ( x + 1) dx =

−2

f −2

f'


2
f + k =
x3 +1 + k
3



∫

cos x
⋅dx
sen 2 x
cos x
⋅dx =
sen 2 x

1
2

−

1
+k
sen xsen −1 x
1
x dx =
+k =−
+k
−1
sen x

^

( x 2 + 1) 5
+k
10

1
1 ( x 2 + 1) 5
( x 2 + 1) 5
2 x ⋅ ( x 2 + 1) 4 dx = ⋅
+k =
+k
2∫ f'
2
5
10
4
f

sen x
⋅ dx
3
x

∫ cos

1
+k
2 cos 2 x

^

sen x
cos −2 x
1
⋅ dx = − ∫ − sen x ⋅ cos −3 x dx = −
+k =
+k
∫ cos 3 x
−2
2 cos 2 x
−3
f'
f
ln ( x + 1)
∫ x + 1 dx

^

ln 2 ( x + 1)
+k
2

ln ( x + 1)
1ln 2 ( x + 1)
dx = ∫ ln ( x + 1) ⋅
dx =
+k
∫ x +1
x +1
2
1
f

∫

cos x
dx
2sen x + 1

f'

2sen x + 1 + k

^
3

1
−
cos x
1
1 (2sen x + 1) 2
dx = ∫ 2 cos x (2sen x + 1) 2 dx = ⋅
+ k = 2sen x + 1 + k
1
2
2
2sen x + 1
f'
−1 / 2
f
2
1

∫

∫

sen 2 x
dx
(1 + cos 2 x ) 2

^

1
+k
2(1 + cos 2 x )

sen 2 x
1
1 (1 + cos 2 x) −1
dx = − ∫ − 2sen 2 x...