A problem in definite integral. What will be the value of $a$ for which the integral 
$$\int \limits^{\infty }_{0}\frac{dx}{a^{2}+(x-\frac{1}{x})^{2}} =\frac{\pi}{5050}$$
where $a^{2}\geq0$
It seems like a standard integral but $x-\frac{1}{x}$ creating all the problem. 
I tried substituting $x-\frac{1}{x}=t$ and then differentiating and squaring and then rearranging but it leads us to nowhere. 
I couldn't find any other substitution to solve this problem. 
Any hint will be of great help. 
Please do not provide the complete solution. 
 A: Thanks Achille Hui for the hint. 
Adding the answer for future reference of users. 
$$\int \limits^{\infty }_{0}\frac{dx}{a^{2}+(x-\frac{1}{x})^{2}} =\frac{\pi}{5050}$$
Hint: $$\int_0^\infty dx = \left(\int_0^1 + \int_1^\infty\right) dx = \int_1^\infty \left(\frac{dx}{x^2} + dx\right) = \int_0^\infty dy\quad\text{ with }\quad  y = x - \frac{1}{x}$$
Change of variable $x_{old} \to \frac{1}{x_{new}}$ for $x_{old} \in (0,1)$.
$$\int \limits^{\infty }_{0}\frac{dx}{a^{2}+(x-\frac{1}{x})^{2}} = \int \limits^{\infty }_{0}\frac{dy}{a^{2}+y^{2}} $$
$$\int \limits^{\infty }_{0}\frac{dy}{a^{2}+y^{2}}= [\frac{1}{a}tan^{-1}\frac{y}{a}]^{\infty }_{0}=\frac{\pi}{2a}$$
Hence,$a=2525$.
A: Just another way: 
Set $x=\frac{e^q}{2}$ ,$dx=\frac{e^q}{2}$
It follows that 
$$
I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{e^q}{a^2+\sinh^2(q)}dq
$$
performing a change of variable $q\rightarrow -q'$ we see that only the numerator changes:
$$
I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{e^{-q'}}{a^2+\sinh^2(q')}dq'
$$
relabelling the dummy variable to $q$ and adding to the original integral yields:
$$
2I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{e^q+e^{-q}}{a^2+\sinh^2(q)}dq=\int_{-\infty}^{\infty}\frac{\cosh(q)}{a^2+\sinh^2(q)}dq=\int_{-\infty}^{\infty}\frac{d\sinh(q)}{a^2+\sinh^2(q)}=\left[\frac{\tanh(\frac{\sinh(q)}{a})}{a}\right]_{-\infty}^{\infty}=\frac{\pi}{a}
$$
And therefore
$$
I=\frac{\pi}{2a}
$$
So $a=2525$
