How do you find the definite integral for: #((2x)/(1+x^2)^(1/3))dx# for the intervals #[0, sqrt7]#?

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Answer 1 · 2017-01-01T23:33:42

First, find the indefinite integral and then evaluate from the lower bound to the upper bound.

#int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = ?#

Let #u = 1 + x^2#, then #du = 2xdx# and:

#int(2x)/(1 + x^2)^(1/3)dx = intu^(-1/3)du#

#int(2x)/(1 + x^2)^(1/3)dx = 3/2u^(2/3) + C#

Reverse the substitution:

#int(2x)/(1 + x^2)^(1/3)dx = 3/2(1 + x^2)^(2/3) + C#

We do not need the constant for the definite integral:

#int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 3/2(1 + x^2)^(2/3)|_0^sqrt(7)#

#int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 3/2(1 + (sqrt(7))^2)^(2/3) - 3/2(1 + 0^2)^(2/3)#

#int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 3/2(8)^(2/3) - 3/2#

#int_0^sqrt(7) (2x)/(1 + x^2)^(1/3)dx = 4.5#