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I have a problem with this integral:

$\int{x^3\over x^8+3}dx$

I try to substitute $x^4$, but I have no idea how continue. Thank you for any help.

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    wolframalpha.com has a great feature that will walk you through the steps of relatively simple integrals like this. Just click "step-by-step solution" once you [have entered the integral](http://www.wolframalpha.com/input/?i=integrate+x%5E3%2F%28x%5E8+%2B+3%29). Unfortunately you have to sign up for a free account to see the steps now.2012-12-05

1 Answers 1

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Let $t=x^4$, then $dt=4x^3\,dx$, and

$\int\frac{x^3}{x^8+3}\,dx=\int\frac{dt}{4(t^2+3)}$

Let $t=\sqrt3u$, then $dt=\sqrt3\,du$, and

$\int\frac{dt}{4(t^2+3)}=\int\frac{\sqrt3\,du}{12(u^2+1)}=\frac{\sqrt3}{12}\int\frac{du}{1+u^2}=\frac{\sqrt3}{12}\arctan u+C=\frac{\sqrt3}{12}\arctan\frac{\sqrt3x^4}{3}+C$