Prove $\int_{0}^{\infty}\frac{2x}{x^8+2x^4+1}dx=\frac{\pi}{4}$

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 $$ \int_{0}^{\infty} \frac{1}{(y^2+1)^2} \ dy=\frac{\pi}{4}.$$

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.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}