Evaluate the integral $int \ x^2(x^3-1)^4 \ dx$ ?

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Answer 1 · 2017-05-03T14:17:39

$int \ x^2(x^3-1)^4 \ dx = 1/15(x^3-1)^5 + C$

We want to find:

$I = int \ x^2(x^3-1)^4 \ dx$

We can perform a simple substitution; Let

$u = x^3-1 => (du)/dx = 3x^2$

If we perform the substitution then we get:

$I = int \ 1/3u^4 \ du$

So we can now integrate to get:

$I = 1/3*1/5u^5 + C$
$\ \ = 1/15 u^5 + C$

And restoring the substitution we get:

$I = 1/15(x^3-1)^5 + C$

Evaluate the integral int \ x^2(x^3-1)^4 \ dx int \ x^2(x^3-1)^4 \ dx ?