I need to prove that for every $p>0$:
$$ \int_0^1\frac{x^{p-1}}{x+1}dx=\int_1^\infty\frac{x^{-p}}{x+1}dx $$
My initial thought is to devide into cases. for $p=1$ the proof is easy because I can calculate the improper integral. Then for $p\neq1$ it gets tricky and seems too complicated. Also, switching sides of the equation doesn't seems to to help much because the intervals are different. Is there a different method for such proof?
As peek-a-boo suggested, you can proceed in the following way through the substitution $\,x=\frac1t\,.$
$\displaystyle\int_0^1\dfrac{x^{p-1}}{x+1}\mathrm dx=\int_{+\infty}^1\dfrac1{t^{p-1}\left(\frac1t+1\right)}\cdot\left(-\dfrac1{t^2}\right)\mathrm dt=$
$=\displaystyle\int_1^{+\infty}\dfrac1{t^p(1+t)}\mathrm dt=\int_1^{+\infty}\dfrac{t^{-p}}{t+1}\mathrm dt\,.$