What is the simplest way to integrate $$I = \int \frac{u^2+1}{(u^2-2u-1)^2}\mathrm{d}u$$ ? I already from Maple know that the answer is $$I = - \frac{u}{u^2 - 2u - 1} + \mathcal{C}$$ and since the answer is so simple, I thought that there must be some clever way to evaluate the integral. I know one could do partial fractions, but why would one want to suffer so much pain ?


Notice that the integral can be rewritten as $$ I = \int \frac{1 + 1/u^2}{(u - 1/u - 2)^2}\mathrm{d}u$$ Now using the substitution $\kappa = u - \frac{1}{u}$, the integral turns into $$ I = \int \frac{\mathrm{d}\kappa}{(\kappa - 2)^2} = -\frac{1}{\kappa-2} = -\frac{u}{u^2-2u-1} + \mathcal{C}$$ as wanted.

Heh, figured this out just after posting the question. This integral had spun around inside my head for quite a few days.

  • 2
    $\begingroup$ But is that the simplest way? =) $\endgroup$ – Pedro Tamaroff Feb 14 '13 at 20:08

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