# How can I evaluate $\int{\frac{x^3}{1+x^8}}\ dx$?

This question came up in a review assignment for my Calculus class, and I've been having difficulty solving it. I don't see what I could substitute for $u$, and it doesn't look like partial fractions are a viable method of doing this problem. Any pointers would be much appreciated.

Evaluate the following indefinite integral: $$\int{\frac{x^3}{1+x^8}}dx$$

Substitute $u=x^4$. The strategy is to get that numerator term $x^3$ to be someone's derivative. (Or a scalar multiple of it.)
• Oh, I see. So we want $x^3$ to be a derivative so that we can use $u$-substitution.The original integral is the same as $\int(x^3)(\frac{1}{1+(x^4)^2})dx$, so we then set $u=x^4$. From there, $du=4x^3dx$ , so I can write the integral as $\frac{1}{4}\int\frac{1}{1+u^2}du$. This simplifies to $\frac14arctan(u)$, and if you plug in $x^4$ for $u$, you end up with the final answer as $\frac14arctan(x^4)$. Did I go wrong anywhere? I'm still unsure that I'd be able to think of substituting like we did here on another problem. – akot717 Aug 29 '17 at 1:19