Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$ $$\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$$
$u=x^4$ $\rightarrow$ $du=4x^3dx$
$x \rightarrow \infty$, $u\rightarrow \infty$
$x\rightarrow 0$, $u\rightarrow 0$
$$=\int_{0}^{\infty}\frac{2x}{u^2+2u+1}\cdot\frac{du}{4x^3}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}{x^2}$$
$$=\frac{1}{2}\int_{0}^{\infty}\frac{1}{u^2+2u+1}\cdot\frac{du}
{\sqrt{u}}$$
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?
 A: By subbing $x=\sqrt{z}$, then $\frac{1}{z^2+1}=u$
$$\int_{0}^{+\infty}\frac{2x}{(x^4+1)^2}\,dx = \int_{0}^{+\infty}\frac{dz}{(z^2+1)^2}=\frac{1}{2}\int_{0}^{1}u^{1/2}(1-u)^{-1/2}\,du$$
that by Euler's beta function and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ equals:
$$ \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\,\Gamma(2)} = \color{red}{\frac{\pi}{4}}$$
as wanted.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Following @Vivek Kaushik helpful answer:
\begin{align}
\color{#f00}{\int_{0}^{\infty}{\dd u \over \pars{u^{2} + 1}^{2}}} & =
-\lim_{\beta \to 1}\totald{}{\beta}\int_{0}^{\infty}{\dd u \over u^{2} + \beta}
= -\lim_{\beta \to 1}\totald{}{\beta}\bracks{{1 \over \root{\beta}}
\int_{0}^{\infty}{\dd u/\!\root{\beta} \over \pars{u/\!\root{\beta}}^{2} + 1}}
\\[3mm] & \stackrel{u/\!\root{\beta}\ \to\ u}{=}\
-\lim_{\beta \to 1}\totald{}{\beta}\pars{\beta^{-1/2}%
\int_{0}^{\infty}{\dd u \over u^{2} + 1}} =
-\pars{-\,\half}\,{\pi \over 2} = \color{#f00}{{\pi \over 4}}
\end{align}
A: Here  is an  approach  using  contour integration  in  case anyone  is
interested. Suppose we seek to verify that
$$\int_0^\infty \frac{2x}{x^8+2x^4+1} dx = \frac{\pi}{4}$$
or alternatively
$$\int_0^\infty \frac{x}{x^8+2x^4+1} dx = \frac{\pi}{8}.$$
We  use a  quarter pizza  slice contour  with the  straight components
$\Gamma_0$ and $\Gamma_1$  on the positive real axis  and the positive
imaginary axis and having radius $R$ ($\Gamma_2.$)
The denominator here is
$$(x^4+1)^2$$
so the poles are double and located at
$$\rho_{0,1,2,3} = \exp(\pi i/4 + 2\pi i k/4)
= \exp(\pi i/4 + \pi i k/2)$$
with $k = 0,1,2,3.$ Fortunately we can see by inspection that only the
first pole $\rho_0$ is inside the contour (argument is $\pi/4.$)
For the residue we get
$$\frac{1}{2\pi i}
\int_{|z-\rho_0|=\epsilon} \frac{z}{z^8+2z^4+1} \; dz.$$
Exploiting the symmetry put $w=z\exp(-\pi i/4)$ and $z=w\exp(\pi i/4)$
to get
$$\exp(\pi i/2) \frac{1}{2\pi i}
\int_{|w\exp(\pi i/4)-1|=\epsilon} 
\frac{w}{w^8-2w^4+1} \; dw
\\ =  \frac{i}{2\pi i}
\int_{|w-1|=\epsilon} 
\frac{w}{w^8-2w^4+1} \; dw.$$
The residue is thus given by
$$i\times \lim_{w\rightarrow 1}
\left(\frac{(w-1)^2 w}{w^8-2w^4+1}\right)'
= i\times \lim_{w\rightarrow 1}
\left(\frac{w}{(w+1)^2 (w^2+1)^2}\right)'
\\ = i\times \lim_{w\rightarrow 1}
\left(\frac{1}{(w+1)^2 (w^2+1)^2}
\\ - \frac{w}{(w+1)^4 (w^2+1)^4} 
(2(w+1)(w^2+1)^2+(w+1)^2 2(w^2+1) 2w\right).$$
This works out to
$$i \times \left(\frac{1}{16} - \frac{16+32}{256}\right)
= -\frac{i}{8}.$$
Returning to the main computation, on  the part of the contour that is
on the positive imaginary axis which is $\Gamma_1$ we obtain
$$\int_{\Gamma_1} \frac{z}{8z^8+2z^4+1} \; dz$$
which yields with $z=\exp(\pi i/2) x$
$$- \int_0^R \frac{\exp(\pi i/2) x}{8x^8+2x^4+1} 
\; \exp(\pi i/2) dx
= \int_{\Gamma_0} \frac{z}{8z^8+2z^4+1} \; dz.$$
Finally we have by the ML bound for the circular component
$$\lim_{R\rightarrow\infty} 
\left|\int_{\Gamma_2} \frac{z}{8z^8+2z^4+1} \; dz\right|
\le \lim_{R\rightarrow\infty} 2\pi R/4 \times 
\frac{R}{8R^8-2R^4+1} = 0.$$
It follows that 
$$\int_0^\infty \frac{x}{8x^8+2x^4+1} \; dx
= \frac{1}{2}\times 2\pi i \times -\frac{i}{8}
= \frac{\pi}{8}$$
which is the claim.
A: 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:
$$I = \int\limits_0^\infty{1+y^2-y^2\over (y^2+1)^2}\,dy = \int\limits_0^\infty{dy\over y^2+1} + {1\over2}\int\limits_0^\infty y\,d{1\over y^2+1}\,dy$$
$$ = \int\limits_0^\infty {dy\over y^2+1} + {1\over2}{y\over y^2+1}\biggr|_0^\infty - {1\over2}\int\limits_0^\infty {dy\over y^2+1} = {1\over2}\arctan y\biggr|_0^\infty = {\pi\over4}.$$
A: Original Solution with Elementary Calculus:
Let $u=x^2, du=2x \ dx.$
Then the integral becomes 
$$\int_{0}^{\infty}\frac{1}{u^4+2u^2+1}\ du.$$
Now notice $u^4+2u^2+1=(u^2+1)^2$, so the integral becomes: 
$$\int_{0}^{\infty}\frac{1}{(u^2+1)^2}\ du.$$
Let $u=\tan(z),du=\sec^2(z)dz.$
So the integral becomes :
$$\int_{0}^{\frac{\pi}{2}}\frac{\sec^2(z)}{\sec^4(z)}\ dz=\int_{0}^{\frac{\pi}{2}}\cos^2(z)\ dz=\int_{0}^{\frac{\pi}{2}}\frac{1+\cos(2z)}{2}\ dz=\frac{\pi}{4}.$$
Another Solution with Fourier Transform
I also was playing around with Fourier Transforms and came across the identity unexpectedly.   
Define the Fourier Cosine Transform, $$\mathcal{F}_{c}(f(x))=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty} f(x) \cos(xy) \ dx.$$ 
Next, consider the inner product defined on Hilbert space $L^2 (0, +\infty),$ the space of square integrable real valued functions over the interval $(0, +\infty),$  defined $$\langle f(x), g(x) \rangle =\int_{0}^{\infty} f(x)g(x) \ dx.$$ It also turns out $\mathcal{F}_{c}$ is a unitary operator, meaning that $\mathcal{F^*}_{c}=\mathcal{F^{-1}}_{c}$   which is the direct result of the Fourier Inversion Theorem.  As a result, we get  $$\langle \mathcal{F}_{c}(f(x)), \mathcal{F}_{c}(g(x)) \rangle=\langle f(x), \mathcal{F^*}_{c}\mathcal{F}_{c}(g(x)) \rangle= \langle f(x), g(x) \rangle.$$ 
Now suppose $f(x)=g(x)=e^{-x}.$ The right hand side gives us 
$$\langle f(x), g(x) \rangle=\int_{0}^{\infty} e^{-2x} \ dx = \frac{1}{2}.$$ On the other hand, $$\langle \mathcal{F}_{c}(f(x)), \mathcal{F}_{c}(g(x)) \rangle= \frac{2}{\pi} \int_{0}^{\infty} \frac{1}{(1+y^2)^2} \ dy.$$ Equating both sides, we get
$$\frac{2}{\pi} \int_{0}^{\infty} \frac{1}{(1+y^2)^2} \ dy=\frac{1}{2},$$ so $$ \int_{0}^{\infty} \frac{1}{(y^2+1)^2} \ dy=\frac{\pi}{4}.$$  
A: Original Solution with Elementary Calculus:
Let
$$ I=\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\int_{0}^{\infty}\frac{2x^{-3}}{x^4+2+x^{-4}}dx. \tag{1}$$
Note that Letting $x\to\frac1x$ gives
$$ I=\int_{0}^{\infty}\frac{\frac2x}{x^{-8}+2x^{-4}+1}\frac{dx}{x^2}=\int_{0}^{\infty}\frac{2x}{x^4+2+x^{-4}}dx. \tag{2}$$
Adding (1) and (2) and changing variable $x^2-x^{-2}\to x$, one has
$$ 2I=\int_0^\infty\frac{2(x+x^{-3})}{x^4+2+x^{-4}}dx=\int_0^\infty\frac{2(x+x^{-3})}{(x^2-x^{-4})^2+4}dx=\int_{-\infty}^\infty\frac{1}{x^2+4}dx=\frac{\pi}{4}$$
and hence
$$ I=\frac{\pi}{8}. $$
