$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is the good substitution to make it integrable?Thanks in advance.

The substitution $t=1/x$ works.  Then, $dx=-\frac{1}{t^2}dt$ and 
$$\begin{align}
I&=\int_0^{\infty}\frac{1}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\
&=\int_0^\infty \frac{1}{a^2+\left(x-\frac1x\right)^2}\frac{1}{x^2}\,dx\\\\
&=\frac12\int_0^\infty \frac{1+\frac{1}{x^2}}{a^2+\left(x-\frac1x\right)^2}\,dx\\\\
&=\frac12\int_0^\infty  \frac{1}{a^2+\left(x-\frac1x\right)^2}\,d\left(x-\frac1x\right)\\\\
&=\left.\frac1{2a} \arctan\left(\frac{x-\frac1x}{a}\right)\right|_{0}^{\infty}\\\\
&\bbox[5px,border:2px solid #C0A000]{=\frac{\pi}{2a}\text{sgn}(a)}
\end{align}$$
If $a>2$, then setting $\frac{\pi}{2a}=\frac{\pi}{5050}$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{a=2525}$$
