Computing $\int (1 - \frac{3}{x^4})\exp(-\frac{x^{2}}{2}) dx$ How does one compute
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
Mathematica gives $(x^{-3} - x^{-1})\exp(-x^2/2)$ which indeed the correct answer, but how does one get there? 
Integration by parts gives $(y + y^{-3})e^{-y^2/2} + \int (y^2 + y^{-2})e^{-y^2/2} dy$, but I'm not sure what to do with the integral.

i tired integration by parts to show to OP but i think it doesn't work
Let's compute this integral :
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
indeed,
$$ u=e^{\dfrac{-x^{2}}{2}},dv = (1 - 3/x^4)\qquad du=-e^{-\dfrac{x^2}{2}}x,\, v=\frac{1}{x^3}+x $$
\begin{align}
\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right)\,dx&=(e^{\dfrac{-x^{2}}{2}} )(\frac{1}{x^3}+x)\biggl|-\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx \\
&=\frac{x^4+1}{e^{\frac{x^2}{2}}x^3}\biggl|-\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx\\
&=\frac{x^4+1}{e^{\frac{x^2}{2}}x^3}\biggl|-\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx
\end{align}
let's calculate $$\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx$$
$$ u=e^{\dfrac{-x^{2}}{2}},dv = \frac{1}{x^3}+x\qquad du=-e^{-\dfrac{x^2}{2}}x,\, v=-\frac{1}{2x^2}+\frac{x^2}{2} $$
\begin{align}
\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx&=-(e^{\dfrac{-x^{2}}{2}})(-\frac{1}{2x^2}+\frac{x^2}{2})\biggl|-\int (-\frac{1}{2x^2}+\frac{x^2}{2})( -e^{-\dfrac{x^2}{2}}x)\,dx\\
&=
\end{align}
