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?

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.
