Calculo samy slideshare
- 1. e.)
∫
2𝑥2+6 𝑥
( 𝑥2+1)(𝑥2+2)2
dx =
𝐴𝑋+𝐵
𝑥2+1
+
𝐶𝑋+𝐷
(𝑋2+2)
+
𝐸𝑋+𝐹
(𝑋2+2)2
2𝑥2 + 6 𝑥 = (𝐴𝑋 + 𝐵) (𝑋2 + 2)2 + (CX+ D) (𝑋2 + 2)(𝑋2 + 2)+ (EX+F)(𝑥2 + 1)
=( 𝐴𝑋 + 𝐵)( 𝑋4 + 4𝑋2 + 4) +(CX +D) ( 𝑋4 + 3𝑋2 + 2)+E𝑋3 + Ex + F𝑥2 + F
= A𝑋5 + 4𝐴𝑋3+ 4AX + B𝑋4 + 4B𝑋2 + 4B + C𝑋5+ 3C𝑋3+ 2CX +D𝑋4+3D𝑋2+ 2D + E𝑋3+ F𝑋2+ EX + F
= (A + C) 𝑋5 + (B +D) 𝑋4+ (4A + 3C +E) 𝑋3 + (4B+ 3D + F) 𝑋2 + (4A + 2CX + E)X+ (4B +F)
A+C= 0 A= -C1)
B+D= 0 B= -D 2)
3) 4D + 3C + E = 0
4) EB + 3D + F = 2
5) 4A + 2C + E = 6
6) 4B + 2D + F = 0
SUSTITUYO 1 EN 3 Y EN 5
-C + E = 0 2C + -2E = 0
-2C + E = 6 -2C + E = 6
E = -6 ; C = -6 ; A=6
SUSTITUYO 2 EN 4 Y 6
-D + F = 2 2D- 2F = -4
-2D + F = 0 -2D + F = 0
-F= -4
F= 4
D = 2
B = -2
∫
6𝑋−2
𝑋2+1
dx + ∫
−6𝑋+2
𝑋2+2
dx + ∫
− 6𝑋+4
(𝑥2+2)2
∫
6𝑋−2
𝑋2+1
dx= ∫
6𝑋
𝑋2+1
dx – 2 ∫
𝑑𝑥
𝑋2+1
U = 𝑥2+1 du= Ex dx
∫
3𝑑𝑈
𝑈
+ 2 tan−1 𝑥 + 𝐶 = 3𝐿𝑛 𝑥2+1 – 2 tan−1 𝑥 + 𝐶
∫
−6𝑋+2
𝑋2+2
= - 6 + ∫
𝑥
𝑋2+2
dx + 2 ∫
𝑑𝑥
𝑋2+2
- 2. = 3Ln𝑋2 + 2 +
2
√2
2 tan−1 𝑥
√2
+ C
∫
− 6𝑋+4
(𝑥2+2)2
= 6 ∫
𝑋
(𝑥2+2)2
𝑑𝑥 + 4 ∫
𝑑𝑥
(𝑥2+2)2
U = 𝑥2 +2 du= 2x dx
= 3 ∫ 𝑢−2 du= -3 𝑢−1 + C= -3 (𝑋2 + 2)−1 + C
X = √2 tan 𝜃 dx = √2𝑠𝑒𝑐2 𝜃d𝜃
√2
4
∫
𝑠𝑒𝑐2 𝜃 d𝜃
𝑠𝑒𝑐4 𝜃
=
√2
4
∫
d𝜃
𝑠𝑒𝑐2 𝜃
=
√2
4
∫ 𝑐𝑜𝑠2 𝜃 d𝜃
=
√2
4
∫
1+cos2𝜃
2
d𝜃 =
√2
8
[∫ 𝑑𝜃 + ∫cos2 𝜃 𝑑𝜃]
=
√2
8
[5 +
𝑠𝑒𝑛2 𝜃
2
] =
√2
8
[tan−1 𝑥
√2
+
2𝑠𝑒𝑛 𝜃 .𝑐𝑜𝑠2𝜃
2
]
=
√2
8
[tan−1 𝑥
√2
+
𝑥
𝑥2+2
.
√2
𝑥2 +2
]
3 Ln | 𝑥2 + 1| – 2 tan−1 𝑥 + 3𝐿𝑛 | 𝑥2 + 2| +
2
√2
tan−1 𝑥
√2
+
√2
8
[tan−1 𝑥
√2
+
√2 𝑥
(𝑥2+2)2
]
D.)
∫ 𝑡𝑎𝑛3 𝑥√ 𝑠𝑒𝑐𝑥 𝑑𝑥
∫
𝑠𝑒𝑛3 𝑥
𝑐𝑜𝑠3 𝑥
√
1
𝑐𝑜𝑠𝑥
𝑑𝑥 = ∫
𝑠𝑒𝑛2 𝑥 . 𝑠𝑒𝑛𝑥 𝑑𝑥
𝑐𝑜𝑠
7
2⁄
𝑥
= ∫
(1 − 𝑐𝑜𝑠2 𝑥) . 𝑠𝑒𝑛𝑥 𝑑𝑥
𝑐𝑜𝑠
7
2⁄
𝑥
Cambiode variable
U=cosx du=-senx dx
∫(1 − 𝑢2). 𝑢
7
2⁄
− 𝑑𝑢 = − ∫ 𝑢
−7
2⁄
+ ∫ 𝑢
−3
2⁄
𝑑𝑢
= −(−
2
5
𝑢−
5
2) + (−2𝑢−
1
2) + 𝐶
=
2
5
(cos 𝑥)−
5
2 − 2 (cos 𝑥 )−
1
2 + C
- 3. c.) ∫
cos𝑥 +𝑥.𝑠𝑒𝑛𝑥−1
( 𝑠𝑒𝑛𝑥−𝑥)2
dx
∫
(cos𝑥−1)
( 𝑠𝑒𝑛𝑥−𝑥)2
𝑑𝑥 + ∫
𝑥.𝑠𝑒𝑛𝑥
( 𝑠𝑒𝑛𝑥−𝑥)2
𝑑𝑥
U= senx – x
du= (cosx – 1)dx
∫
𝑑𝑢
𝑢2
= - 𝑢−1 = -
1
𝑠𝑒𝑛𝑥−𝑥
+ C
∫
𝑥 .𝑠𝑒𝑛𝑥
( 𝑠𝑒𝑛𝑥−𝑥)
dx =
𝑐𝑜𝑠𝑥+1
𝑠𝑒𝑛𝑥−𝑥
∫
cos 𝑥 + 𝑠𝑒𝑛𝑥 − 1
(𝑠𝑒𝑛𝑥 − 𝑥)2 =
cos 𝑥 + 1
𝑠𝑒𝑛𝑥− 𝑥
−
1
𝑠𝑒𝑛𝑥 − 𝑥
=
cos𝑥
𝑠𝑒𝑛𝑥−𝑥
+ 𝐶
b.)
= ln(x + √1 + 𝑥2) x – ∫
1
√𝑥2+1
𝑥 𝑑𝑥
= x ln (√𝑥2 + 1 + x) - ∫
𝑥
√𝑥2+1
𝑑𝑥
∫
𝑥
√𝑥2+1
dx = √𝑥2 + 1
= x ln (√𝑥2 + 1 + 𝑥) - √𝑥2 + 1
= x ln (√𝑥2 + 1 + 𝑥) – √𝑥2 + 1 + 𝐶
a.)
∫
1
4𝑢 √ 𝑢 − 4
𝑑𝑢
1
4
∫
1
𝑢 √ 𝑢−4
1
4
∫
2
𝑣2+4
dv
1
4
2∫
1
𝑣2 +4
dv
1
4
2∫
1
2𝑤2+2
dw
1
4
2∫
1
2
1
𝑤2+1
dw
1
4
2
1
2
∫
1
𝑤2+1
dw
1
4
2
1
2
tan−1 𝑤
![= 3Ln𝑋2 + 2 +
2
√2
2 tan−1 𝑥
√2
+ C
∫
− 6𝑋+4
(𝑥2+2)2
= 6 ∫
𝑋
(𝑥2+2)2
𝑑𝑥 + 4 ∫
𝑑𝑥
(𝑥2+2)2
U = 𝑥2 +2 du= 2x dx
= 3 ∫ 𝑢−2 du= -3 𝑢−1 + C= -3 (𝑋2 + 2)−1 + C
X = √2 tan 𝜃 dx = √2𝑠𝑒𝑐2 𝜃d𝜃
√2
4
∫
𝑠𝑒𝑐2 𝜃 d𝜃
𝑠𝑒𝑐4 𝜃
=
√2
4
∫
d𝜃
𝑠𝑒𝑐2 𝜃
=
√2
4
∫ 𝑐𝑜𝑠2 𝜃 d𝜃
=
√2
4
∫
1+cos2𝜃
2
d𝜃 =
√2
8
[∫ 𝑑𝜃 + ∫cos2 𝜃 𝑑𝜃]
=
√2
8
[5 +
𝑠𝑒𝑛2 𝜃
2
] =
√2
8
[tan−1 𝑥
√2
+
2𝑠𝑒𝑛 𝜃 .𝑐𝑜𝑠2𝜃
2
]
=
√2
8
[tan−1 𝑥
√2
+
𝑥
𝑥2+2
.
√2
𝑥2 +2
]
3 Ln | 𝑥2 + 1| – 2 tan−1 𝑥 + 3𝐿𝑛 | 𝑥2 + 2| +
2
√2
tan−1 𝑥
√2
+
√2
8
[tan−1 𝑥
√2
+
√2 𝑥
(𝑥2+2)2
]
D.)
∫ 𝑡𝑎𝑛3 𝑥√ 𝑠𝑒𝑐𝑥 𝑑𝑥
∫
𝑠𝑒𝑛3 𝑥
𝑐𝑜𝑠3 𝑥
√
1
𝑐𝑜𝑠𝑥
𝑑𝑥 = ∫
𝑠𝑒𝑛2 𝑥 . 𝑠𝑒𝑛𝑥 𝑑𝑥
𝑐𝑜𝑠
7
2⁄
𝑥
= ∫
(1 − 𝑐𝑜𝑠2 𝑥) . 𝑠𝑒𝑛𝑥 𝑑𝑥
𝑐𝑜𝑠
7
2⁄
𝑥
Cambiode variable
U=cosx du=-senx dx
∫(1 − 𝑢2). 𝑢
7
2⁄
− 𝑑𝑢 = − ∫ 𝑢
−7
2⁄
+ ∫ 𝑢
−3
2⁄
𝑑𝑢
= −(−
2
5
𝑢−
5
2) + (−2𝑢−
1
2) + 𝐶
=
2
5
(cos 𝑥)−
5
2 − 2 (cos 𝑥 )−
1
2 + C](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)