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#