Is there a quicker way of doing this integral? I have an integral $\int_0^\infty {x^2\over 1+x^4} dx$. I  gave it a go and it turned out quite messy, so I consulted Wolfram Alpha but the steps given there seem rather long winded too. Is there is a faster way of doing the integral? 
 A: $$\int_0^\infty\frac{x^2}{1+x^4}dx$$
Let, $t=\frac{1}{x}$, and, $dt=\frac{-dx}{x^2}$
$$\int_\infty^0 \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$
$$=\int_\infty^0\frac{-dt}{1+t^4} =\int_0^\infty \frac{dt}{1+t^4}$$
Follow Norberts solution after that.

Another way to do this is
$$
I=\int_0^\infty\frac{x^2}{1+x^4}dx
$$
Let, $x= \sqrt{\tan\theta}$, then $dx=\frac{1}{2\sqrt{\tan\theta}}\sec^2\theta d\theta$
$$
I=\int_0^{\frac{\pi}{2}} \frac{\tan\theta}{1+\tan^2\theta}\times\frac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$$
$$
I=\frac{1}{2}\times\int_0^{\frac{\pi}{2}}\sqrt{\tan\theta}
$$
also,
$
I=\frac{1}{2}\times\int_0^{\frac{\pi}{2}}\sqrt{\cot\theta}
$
hence,
$$
4I=\int_0^{\frac{\pi}{2}}\sqrt{\tan\theta}+\sqrt{\cot\theta}
$$
$$
4I=\int_0^{\frac{\pi}{2}} \frac{\sin\theta + \cos\theta}{\sqrt{\sin\theta\cos\theta}}
$$
$$=\sqrt2 \int_0^{\frac{\pi}{2}} \frac{(\sin\theta + \cos\theta)}{\sqrt{1-(\sin\theta - \cos\theta)^2}}$$
Let $t=\sin\theta - \cos\theta$, then 
$$4I=\sqrt2 \int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}}$$
$$I=\frac{1}{2\sqrt2}\left(\sin^{-1}(1)-\sin^{-1}(-1)\right) =\frac{\pi}{2\sqrt2}$$
:)
$$ \int_0^1 \left(\sqrt[3]{1-x^7}  \right)$$
$$\frac{a}{d}$$
A: As SauravTomar pointed out
$$
\int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=\int\limits_{0}^\infty\frac{1}{x^4+1}dx
$$
so
$$
\int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=
\frac{1}{2}\int\limits_{0}^\infty\frac{x^2+1}{x^4+1}dx=
\frac{1}{2}\int\limits_{0}^\infty\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx
$$
$$
=\frac{1}{2}\int\limits_{0}^\infty\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}dx=
\frac{1}{2}\int\limits_{-\infty}^\infty\frac{dt}{t^2+2}dx=
\frac{\pi}{2\sqrt{2}}.
$$
