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

Note
\begin{eqnarray}
I&=&\int_{0}^{\infty}\frac{1}{a^2+\left(x-\frac{1}{x}\right)^2}dx\\
&=&\int_{0}^{\infty}\frac{1}{x^2+\frac{1}{x^2}+(a^2-2)}dx\\
&=&\int_{0}^{\infty}\frac{x^2}{x^4+kx^2+1}dx.
\end{eqnarray}
Here $k=a^2-2$.
Suppose $a>2$. Letting $x^4+kx^2+1=(x^2+r_1^2)(x^2+r_2^2)$ where $r_1,r_2>0$, we have
\begin{eqnarray}
I&=&\frac1{r_1^2-r_2^2}\left(r_1^2\int_{0}^{\infty}\frac{1}{x^2+r_1^2}-r_2^2\int_{0}^{\infty}\frac{1}{x^2+r_2^2}\right)\\
&=&\frac1{r_1^2-r_2^2}\frac{\pi(r_1-r_2)}{2}\\
&=&\frac{\pi}{2}\frac1{r_1+r_2}.
\end{eqnarray}
Noting $r_1^2+r_2^2=a^2-2$ and $r_1^2r_2^2=1$, we have $r_1+r_2=\sqrt{r_1^2+r_2^2+2r_1r_2}=a$. So
$$ I=\frac{\pi}{2a}.$$
Letting $\frac{\pi}{2a}=\frac{\pi}{5050}$ gives $a=2525$.
