<s>[INST] 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.
 [/INST] "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.
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