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I'm trying to evaluate the following indefinite integral:

$$ \int\frac{e^{-x^2}}{(1+2x^2)^2}dx $$

According to Wolfram|Alpha, this integral evaluates to:

$$ \int \frac{e^{-x^2}}{(1+2 x^2)^2} dx = \frac{1}{4}\left(\sqrt{\pi} \mathrm{erf}(x)+\frac{2x e^{-x^2}}{(2 x^2+1)}\right)+\mathrm{constant} $$

Is there a quick way of evaluating this integral to get that answer? A fairly obvious-looking integration-by-parts approach didn't work but looked fairly promising:

\begin{align} \int \frac{e^{-x^2}}{(1+2 x^2)^2} dx &= \frac{1}{4}\int \frac{e^{-x^2}}{x}\times\frac{4x}{(1+2 x^2)^2}dx \\ &=\frac{1}{4}\left(\frac{-e^{-x^2}}{x(1+2x^2)}+\int\frac{1}{(1+2x^2)}\times\frac{-2x^2e^{-x^2}-e^{-x^2}}{x^2}dx\right) \\ &=\frac{1}{4}\left(-\frac{e^{-x^2}}{x(1+2x^2)}-\int\frac{e^{-x^2}}{x^2}dx\right) \end{align}

If both the boundary term and the integrand were multiplied by $-2x^2$, this would be exactly what we want, since the integral on the right would then evaluate to $\sqrt{\pi}\mathrm{erf}(x)$. Is there some other integration by parts that I've missed? Or is it only possible to do the integral by more complicated means?

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  • $\begingroup$ if the regular methods don't work then my next port of call would be differentiating under the integral or contour integration $\endgroup$ – user27182 Apr 25 '13 at 23:05
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Try doing integration by parts on this term as well with $u = \exp(-x^2)$ and $dv = \frac{1}{x^2}dx$. Combine some terms and I think you'll have your result.

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  • $\begingroup$ Thanks for you help. I'll let you know how it goes! $\endgroup$ – John Gowers Apr 25 '13 at 23:07
  • $\begingroup$ I edited my answer because while you dropped a negative in your LaTeX code, the second term should be negative in the last equality. $\endgroup$ – Cameron Williams Apr 25 '13 at 23:27
  • $\begingroup$ Yes - thanks. I'll change the incorrect $-$ sign now. Also, this method did indeed work! $\endgroup$ – John Gowers Apr 26 '13 at 10:19

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