Hola, necesito un primitiva para estas funciones:
Traten, en lo posible, de no usar fracciones parciales.
1)- 1/(x(x^2 +1))
2)- 1/(x+√x)
3)- √(1+√x)
Gracias!!
Hola
1)-
ʃ dx /(x (x^2 +1)) = ʃ (x^2 + 1 - x^2) dx /(x (x^2 +1))
ʃ dx /(x (x^2 +1)) = ʃ (x^2 + 1)dx /(x (x^2 +1)) - ʃ(x^2) dx /(x (x^2 +1))
ʃ dx /(x (x^2 +1)) = ʃ dx/x - ʃ(x dx) /(x^2 +1)
ʃ dx /(x (x^2 +1)) = ʃ dx/x - (1/2) ʃ(2 x dx) /(x^2 +1)
ʃ dx /(x (x^2 +1)) = ʃ dx/x - (1/2) ʃd(x^2 + 1) /(x^2 +1)
ʃ dx /(x (x^2 +1)) = ln(x) - (1/2) ln(x^2 +1) + C
***********
2)-
ʃ dx / (x + √x)
u = √x
x = u^2
dx = 2 u du
= ʃ 2 u du / (u^2 + u)
= 2 ʃ du / (u + 1)
= 2 ln(u + 1) + K
= 2 ln(√(x) + 1) + K
************
3)-
ʃ √(1+√x) dx
u = 1 + √x
x = (u - 1)^2
dx = 2 (u - 1) du
ʃ √(1+√x) dx = ʃ √(u) 2 (u - 1) du
ʃ √(1+√x) dx = 2 ʃ (u^(3/2) - u^(1/2)) du
ʃ √(1+√x) dx = 2 ( (2/5) u^(5/2) - (2/3) u^(3/2)) + K
ʃ √(1+√x) dx = (4/5) (1 + √x)^(5/2) - (4/3) (1 + √x)^(3/2) + K
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Answers & Comments
Hola
1)-
ʃ dx /(x (x^2 +1)) = ʃ (x^2 + 1 - x^2) dx /(x (x^2 +1))
ʃ dx /(x (x^2 +1)) = ʃ (x^2 + 1)dx /(x (x^2 +1)) - ʃ(x^2) dx /(x (x^2 +1))
ʃ dx /(x (x^2 +1)) = ʃ dx/x - ʃ(x dx) /(x^2 +1)
ʃ dx /(x (x^2 +1)) = ʃ dx/x - (1/2) ʃ(2 x dx) /(x^2 +1)
ʃ dx /(x (x^2 +1)) = ʃ dx/x - (1/2) ʃd(x^2 + 1) /(x^2 +1)
ʃ dx /(x (x^2 +1)) = ln(x) - (1/2) ln(x^2 +1) + C
***********
2)-
ʃ dx / (x + √x)
u = √x
x = u^2
dx = 2 u du
ʃ dx / (x + √x)
= ʃ 2 u du / (u^2 + u)
= 2 ʃ du / (u + 1)
= 2 ln(u + 1) + K
= 2 ln(√(x) + 1) + K
************
3)-
ʃ √(1+√x) dx
u = 1 + √x
x = (u - 1)^2
dx = 2 (u - 1) du
ʃ √(1+√x) dx = ʃ √(u) 2 (u - 1) du
ʃ √(1+√x) dx = 2 ʃ (u^(3/2) - u^(1/2)) du
ʃ √(1+√x) dx = 2 ( (2/5) u^(5/2) - (2/3) u^(3/2)) + K
ʃ √(1+√x) dx = (4/5) (1 + √x)^(5/2) - (4/3) (1 + √x)^(3/2) + K
*******************