My probability textbook has a solution that turns a certain density function into something more familiar. Starting with,

$\int\_{-\infty}^{\infty} \frac{dx}{(1+x^2)^m}$

We substitute in:

$v = (1 + x^2)^{-1}$

to get:

$\int\_{-\infty}^{\infty} \frac{dx}{(1+x^2)^m}$ = $\int\_{0}^{1} v^{m-\frac{3}{2}}(1-v)^{-\frac{1}{2}} dv$

I'm not quite sure how they got from the left side of that to the right...