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

"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.
