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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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Have you tried using $\tan^{-1}$? –  Ben Dec 5 '12 at 1:49
    
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. Unfortunately you have to sign up for a free account to see the steps now. –  orlandpm Dec 5 '12 at 1:55

1 Answer 1

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

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

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