TITLE: Solve${{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$
QUESTION [5 upvotes]: Please help me to evaluate the integral:
$\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$ 
Thanks.

REPLY [4 votes]: With $x = u^{1/3}$ we get
$$
\int\left(x^{4} \over 1 + x^{6}\right)^{2}\,{\rm d}x
=
{1 \over 3}\int{u^{2} \over \left(1 + u^{2}\right)^{2}}\,{\rm d}u
$$
Set $u = \tan\left(\theta\right)$
\begin{align}
{1 \over 3}\int{\tan^{2}\left(\theta\right) \over \sec^{4}\left(\theta\right)}\,
\sec^{2}\left(\theta\right)\,{\rm d}\theta
&=
{1 \over 3}\int\sin^{2}\left(\theta\right)\,{\rm d}\theta
=
{1 \over 3}\int{1 - \cos\left(2\theta\right) \over 2}\,{\rm d}\theta
\\&=
{1 \over 6}\,\theta - {1 \over 12}\,\sin\left(2\theta\right)
=
{1 \over 6}\,\theta
-
{1 \over 6}\,{\tan\left(\theta\right)\over \tan^{2}\left(\theta\right) + 1}
\\&=
{1 \over 6}\left\lbrack
\arctan\left(u\right) - {u \over u^{2} + 1}
\right\rbrack
=
{1 \over 6}\left\lbrack
\arctan\left(x^{3}\right) - {x^{3} \over x^{6} + 1}
\right\rbrack
\end{align}