Integration of $\frac{e^{-x^2/4}}{(2+x^2)^2}$ I was doing a question where I needed to do this integration
$$\int\frac{e^{-x^2/4}}{(2+x^2)^2}dx$$
This answer on Wolfram alpha agrees with the answer I expect to get. However I can't see how to do this integration (and I do not have Wolfram Pro, so cannot see any step-by-step solution).
How can I do this integral? 

Write $$f(x) = \frac{e^{-x^2/4}}{(2+x^2)^2} = \frac{2x}{(2+x^2)^2} \cdot \frac{e^{-x^2/4}}{2x}$$ and integrate by parts with the choice $$u = \frac{e^{-x^2/4}}{2x}, \quad du = -e^{-x^2/4}\frac{(2+x^2)}{4x^2} \, dx, \\ dv =  \frac{2x}{(2+x^2)^2} \, dx, \quad v = -\frac{1}{2+x^2}.$$  Hence $$\int f(x) \, dx = -\frac{e^{-x^2/4}}{2x(2+x^2)} - \int \frac{e^{-x^2/4}}{4x^2} \, dx.$$  We again perform another integration by parts with the choice $$u = e^{-x^2/4}, \quad du = -\frac{x}{2} e^{-x^2/4} \, dx, \\ dv = \frac{1}{4x^2} \, dx, \quad v = -\frac{1}{4x},$$ to obtain $$\int f(x) \, dx = -\frac{e^{-x^2/4}}{2x(2+x^2)} + \frac{e^{-x^2/4}}{4x} + \frac{1}{8} \int e^{-x^2/4} \, dx.$$  This last integral does not have an elementary closed-form antiderivative but it is related to the familiar "error function", thus $$f(x) = \frac{x e^{-x^2/4}}{4(2+x^2)} + \frac{\sqrt{\pi}}{8}\operatorname{erf}(x/2) + C.$$
