How do you find the indefinite integral of $int x(1-3x^2)^4dx$ ?

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Answer 1 · 2017-03-03T18:41:39

Use substitution.

$intx(1-3x^2)^4dx$

Set $u=1-3x^2$ . Then $du=-6xdx$ and $dx=-1/6du$ . Substituting into the integral:

$=>-1/6intu^4du$

This is now a basic integral.

We get $-1/6*(1/5u^5)+C$ .

Substituting back in for $u$ and simplifying:

$=>-1/30(1-3x^2)^5+C$

How do you find the indefinite integral of int x(1-3x^2)^4dxint x(1-3x^2)^4dx?