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In Apostol, the case that the partial fraction is presented as $\frac{C}{(u^2 + a^2)^m}$ can be done with the following reduction formula:

$$\int \frac{du}{(u^2 + a^2)^m} = \frac{1}{2a^2(m-1)} \frac{u}{(u^2 + a^2)^{m-1}} + \frac{2m - 3}{2a^2(m-1)} \int \frac{du}{(u^2 + a^2)^{m-1}}$$

The author states that this can be shown using integration by parts. I cannot figure how - i tried taking $dv = (u^2 + a^2)^{m-1}$ or $dv = 1 \Rightarrow v = u$. In case the $m = 2$, the integral can be done with trigonometric substitution. However, this general case is not as clear. Can someone show how this is obtained with integration by parts?

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I suggest that you write $$ \frac{1}{(u^2+a^2)^{m-1}} = u\cdot\frac{u}{(u^2+a^2)^m}+\frac{a^2}{(u^2+a^2)^m}, $$ and integrate by parts in the first term in the right-hand side, and then rearrange your terms.

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    $\begingroup$ Cool, thank you! Got it! @mickep, how did you come up with this transformation? Did you know it beforehand, solved similar examples, or you just see the pattern somehow? I think I myself would have never found how to integrate this, but with your transformation it is so simple! $\endgroup$ – John Aug 13 at 20:17
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    $\begingroup$ @John Glad it helped. I've seen and done similar manipulations before. $\endgroup$ – mickep Aug 14 at 5:02

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