Evaluate $I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1}$ 
$$I(x)=\int_0^\infty\frac{1}{x^2t^2}\frac{dt}{(\frac{c}{xt}-\frac{mt}{x})^2+1},$$

I used
$$u=\frac{c}{xt}-\frac{mt}{x}$$
and
$$du=-\left(\frac{c}{xt^2}+\frac{m}{x}\right)dt$$
but I'm not able to find the limits at $t=0$ and $t=\infty$. Neither are straight forward to use as
$$\int_0^\infty\frac{1}{1+t^2}dt=\pi/2$$
Any hints on how to proceed?
 A: Rearrange the integral as follows
\begin{align}
& \int_0^\infty\frac{1}{x^2t^2}\frac{1}{(\frac{c}{xt}-\frac{mt}{x})^2+1}dt \\
=& \frac1{2cx} \int_0^\infty \frac{(\frac{c}{xt^2}+\frac mx)+ (\frac{c}{xt^2}-\frac mx )}{(\frac{c}{xt}-\frac{mt}{x})^2+1}dt\\
=& \frac1{2cx} \int_0^\infty \frac{d(\frac {mt }x-\frac{c}{xt})}{(\frac{c}{xt}-\frac{mt}{x})^2+1}
 -\frac1{2cx} \int_0^\infty \frac{d(\frac{c}{xt}+\frac {mt }x )}{(\frac{c}{xt}+\frac{mt}{x})^2+1-\frac{4cm}{x^2}}\\
\end{align}
where the first integral is of the form $\int_0^\infty\frac{du}{1+u^2}$ and the second integral vanishes.
A: Under the interchange $t\leftrightarrow \frac{1}{t}$ we have that
$$I = \int_{0}^\infty \frac{dt}{c^2\left(t-\frac{m}{ct}\right)^2+ x^2}$$
The integral is even and converges so we can say
$$I = \frac{1}{2}\int_{-\infty}^\infty \frac{dt}{c^2\left(t-\frac{m}{ct}\right)^2 + x^2}$$
Lastly we'll use the theorem
$$p.v.\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)dx = p.v.\int_{-\infty}^\infty f(x)dx$$
for $a>0$ to get
$$I = \frac{1}{2}\int_{-\infty}^\infty \frac{dt}{c^2t^2+x^2} = \frac{\pi}{2|xc|}$$
assuming $mc>0$
