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?

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 $$ \int\limits_{0}^\infty\frac{x^2}{x^4+1}dx= \frac{1}{2}\int\limits_{0}^\infty\frac{x^2+1}{x^4+1}dx= \frac{1}{2}\int\limits_{0}^\infty\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx $$ $$ =\frac{1}{2}\int\limits_{0}^\infty\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}dx= \frac{1}{2}\int\limits_{-\infty}^\infty\frac{dt}{t^2+2}dx= \frac{\pi}{2\sqrt{2}}. $$