Evaluate $\int \frac{dx}{(x^2 + 1)^\alpha}$

Question

I couldn't follow a step while reading this answer. Since I do not have enough reputation to post this as a comment, I'm asking a question instead. The answer uses "partial integration" to write this $$ \int \frac{dv}{(v^2 + 1)^\alpha} = \frac{v}{2(\alpha-1)(v^2+ 1)^{\alpha - 1}} + \frac{2\alpha -3}{2\alpha - 2}\int \frac{dv}{(v^2 + 1)^{\alpha -1}} $$ I would like to know what this technique is, and how this equality follows from it.

Answer 1

"Partial integration" just means integration by parts. The important step here is the writing of the fraction as $$ \frac{1}{(v^2+1)^{\alpha}} = \frac{1}{(v^2+1)^{\alpha}} - \frac{1}{(v^2+1)^{\alpha-1}} + \frac{1}{(v^2+1)^{\alpha-1}} \\ = -\frac{v^2}{(v^2+1)^{\alpha}} + \frac{1}{(v^2+1)^{\alpha-1}}. $$ Then you integrate the first term by parts, integrating $\frac{v}{(v^2+1)^{\alpha}}$ and differentiating $v$; this gives $$ \int \left( \frac{1}{(v^2+1)^{\alpha}}-\frac{1}{(v^2+1)^{\alpha-1}} \right) \, dv = -\int \frac{v^2}{(v^2+1)^{\alpha}} \, dv \\ = \frac{v}{2(\alpha-1)(v^2+1)^{\alpha-1}} - \frac{1}{2(\alpha-1)}\int \frac{dv}{(v^2+1)^{\alpha-1}}, $$ and then rearranging gives the result.

Answer 2

Try integration by parts $$f= (v^2+1)^{-\alpha} ; g' =1 \\ f'=(-\alpha) (v^2+1)^{-\alpha-1}(2v) ; g=v$$

P.S. After integration by Parts, you write the $v^2$ in the numerator as $$v^2=(v^2+1)-1$$ and split the integral in two.