Integral $\int \exp\left(-\frac{z^2}{2}\right) \frac{z^4+2z^2-1}{(z^2+1)^2}dz$

Question

Integrate $$\int \exp\left(-\frac{z^2}{2}\right) \frac{z^4+2z^2-1}{(z^2+1)^2}dz$$

This function has exponential factor and rational factor.

I know already that the solution is

$$-\frac{z\exp\left(-\frac{z^2}{2}\right)}{z^2+1} + C$$

But I have no idea on what method should I use to find it.

Answer 1

$$I=\int e^{\large-\frac{z^2}{2}}\, \frac{z^4+2z^2-1}{(z^2+1)^2}dz$$ With some algebra we have: $$z^4+z^2+z^2-1=z^2(z^2+1) +z^2-1$$ $$\Rightarrow \frac{z^4+2z^2-1}{(z^2+1)^2}=\frac{z^2}{z^2+1}+\frac{z^2-1}{(z^2+1)^2}$$ $$\Rightarrow I= \color{blue}{\int e^{-\frac{z^2}{2}}\frac{z^2}{z^2+1}dz}+\color{red}{\int e^{-\frac{z^2}{2}} \frac{z^2-1}{(z^2+1)^2}dz}$$ Now since we have similarly to here:$$\int \frac{z^2-1}{(z^2+1)^2}dz=-\frac{z}{z^2+1}+C$$ Integrating by parts the second integral gives us: $$\int e^{-\frac{z^2}{2}} \, \frac{z^2-1}{(z^2+1)^2}dz=\color{red}{-e^{-\frac{z^2}{2}}\frac{z}{z^2+1}-\int e^{-\frac{z^2}{2}}\frac{z^2}{z^2+1}dz}$$ $$\require{cancel}\Rightarrow I= \cancel{\color{blue}{\int e^{-\frac{z^2}{2}}\frac{z^2}{z^2+1}dz}}\color{red}{-e^{-\frac{z^2}{2}}\frac{z}{z^2+1}-\cancel{\int e^{-\frac{z^2}{2}}\frac{z^2}{z^2+1}dz}}$$ $$\Rightarrow \int e^{\large-\frac{z^2}{2}}\, \frac{z^4+2z^2-1}{(z^2+1)^2}dz=-e^{-\frac{z^2}{2}}\frac{z}{z^2+1}+C$$

Answer 2

Here is another method that relies on a small trick.

If $f$ and $g$ are differentiable functions, as $$\frac{d}{dx} (e^{f(x)} g(x)) = e^{f(x)}[f'(x) g(x) + g'(x)],$$ it is immediate that $$\int e^{f(x)} \left (f'(x) g(x) + g'(x) \right ) \, dx = e^{f(x)} g(x) + C.$$

In the case of the integral $$\int e^{-\frac{x^2}{2}} \frac{x^4 + 2x^2 - 1}{(x^2 + 1)^2}, dx,$$ Here $f(x) = -x^2/2$, $f'(x) = -x$. For the function $g$ we have $$f'(x) g(x) + g'(x) = - x g(x) + g'(x) = \frac{x^4 + 2x^2 - 1}{(x^2 + 1)^2}.$$

Conjuring up the function $g(x)$ is the hard bit. The squared term appearing in the denominator suggests we try something like: $g(x) = h(x)/(x^2 + 1)$. After a little trial and error it is not too hard to see that $h(x)$ must be $-x$ and we have $$g(x) = -\frac{x}{x^2 + 1}.$$ Thus $$\int e^{-\frac{x^2}{2}} \frac{x^4 + 2x^2 - 1}{(x^2 + 1)^2}, dx = -\frac{x}{x^2 + 1} e^{-\frac{x^2}{2}} + C.$$