$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ 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.

It works, indeed. If we set $x=\frac{1}{2}\left(y+\sqrt{4+y^2}\right)$ we get:
$$ I = \int_{0}^{+\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2}\left(1+\frac{y}{\sqrt{y^2+4}}\right) $$
hence:
$$ I = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{dy}{y^2+a^2} = \frac{\pi}{2a}$$
since the odd part of the previous integrand function clearly vanishes by symmetry.
