Hello,

Are the following two integrals equal?

$2\displaystyle\int_0^\infty\frac{t^a dt}{((b+1)t+(b-1))^2}=\frac{2}{b^2-1}\bigg(\frac{b-1}{b+1}\bigg)^{a+1}\displaystyle\int_0^\infty\frac {u^adu}{(u+1)^2}$ where $u=\frac{(b+1)t}{b-1}$

If they are equal, how are they equal explaining the denominator term $b^2-1$ on the right hand side integral.