Solve${{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$ Please help me to evaluate the integral:
$\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$ 
Thanks.
 A: Notice that:
$$x^8 = (x^6 + 1)x^2 - x^2$$
So that your integrand takes the form:
$$\int\frac{x^2}{1+x^6}dx - \int\frac{x^2}{(1+x^6)^2}dx$$
Now substitute $u = x^3, du = 3 x^2 dx$:
$$\int\frac{1}{3(1+u^2)}du - \int\frac{1}{3(1+u^2)^2}du$$
The first integral is a multiple of $\arctan(u)$, and the second can be solved dividing again into two parts:
$$\int\frac{1}{3(1+u^2)^2}du = \int\frac{1}{3(1+u^2)}du - \int\frac{u^2}{3(1+u^2)^2}du$$
I think you get the idea..
A: 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}
A: Try substitutions of the form $u = x^k$, where $k$ should divide the exponent 6, and work on each of them. At least one of these will look much more familiar. Then work on that one, using the usual tricks.  
A: Here is a result by Maple for indefinite integral. I managed to get the more compact form 
$$ \displaystyle {{\int{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx={\frac {\arctan \left( {x}^{3} \right) {x}^{6}+\arctan \left( {x}^{3}\right) -{x}^{3}}{6\,{x}^{6}+6}}$$
The final result of the definite integral by maple is
$$-\frac{1}{12}+\frac{\pi}{8}  $$
