TITLE: Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$
QUESTION [10 upvotes]: $$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$
$u=x^4$ $\rightarrow$ $du=4x^3dx$
$x \rightarrow \infty$, $u\rightarrow \infty$
$x\rightarrow 0$, $u\rightarrow 0$
$$=\int_{0}^{\infty}\frac{2x}{u^2+2u+1}\cdot\frac{du}{4x^3}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}{x^2}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}
{\sqrt{u}}$$
Convert to partial fractions
$$\int_{0}^{\infty}\frac{1}{\sqrt{u}}-\frac{2}{u+1}+\frac{1}{(u+1)^2}du$$
$$=\left.2\sqrt{u}\right|_{0}^{\infty}-\left.2\ln(1+x)\right|_{0}^{\infty} -\left.\frac{1}{1+u}\right|_{0}^{\infty}$$
Where did I went wrong during my calculation?

REPLY [2 votes]: At first - substitution:
$$I=\int\limits_0^\infty{2x\,dx\over x^8+2x^4+1} = \int\limits_0^\infty{d(x^2)\over x^8+2x^4+1} = \int\limits_0^\infty{dy\over (y^2+1)^2}.$$
And then - by parts:
$$I = \int\limits_0^\infty{1+y^2-y^2\over (y^2+1)^2}\,dy = \int\limits_0^\infty{dy\over y^2+1} + {1\over2}\int\limits_0^\infty y\,d{1\over y^2+1}\,dy$$
$$ = \int\limits_0^\infty {dy\over y^2+1} + {1\over2}{y\over y^2+1}\biggr|_0^\infty - {1\over2}\int\limits_0^\infty {dy\over y^2+1} = {1\over2}\arctan y\biggr|_0^\infty = {\pi\over4}.$$