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.
On
The reason your integration by parts is not working good directly here, is that you do not want to let the constant $1$ and the $3/x^4$ take the same role in the integration by parts (i.e. raise or lower both of them), since then you will just keep going. It is better to "move them towards each other":
For one part: $$ \int -\frac{3}{x^4}e^{-x^2/2}\,dx=\frac{1}{x^3}e^{-x^2/2}+\int \frac{1}{x^3}xe^{-x^2/2}\,dx $$ For the other part, we don't see a primitive of $e^{-x^2/2}$, so we write the $1$ as $1=\frac{1}{x}x$, and then integrate by parts $$ \begin{aligned} \int e^{-x^2/2}\,dx &= \int \frac{1}{x} xe^{-x^2/2}\,dx\\ &=-\frac{1}{x}e^{-x^2/2}-\int\frac{1}{x^2}e^{-x^2/2}\,dx \end{aligned} $$
The "funny" thing now is that the integrals that are left just cancel (this is lucky, and the exercise is made so that this should happen, in the general case one would meet error functions). Adding, we find that $$ \int \Bigl(1-\frac{3}{x^4}\Bigr)e^{-x^2/2}\,dx=\Bigl(\frac{1}{x^3}-\frac{1}{x}\Bigr)e^{-x^2/2}+C. $$
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}