# How to do $\int e^\frac{-1}{2x^2}x^{-5}dx$?

How to do $\int e^\frac{-1}{2x^2}x^{-5}dx$?

The original problem solve $xy'+\frac{y}{x^2}=\frac{1}{x^4}$

I simplified and found the integrating factor $e^\frac{-1}{2x^2}$ now the RHS of the equation is $\int e^\frac{-1}{2x^2}x^{-5}dx$

I have tried integration by parts with $u=e^\frac{-1}{2x^2}$ and $v=x^{-5}$ but this only seems to make it worse.

• Try doing a $u$ substitution with $u=1/x$. – Sean Lake Oct 10 '17 at 3:26
• $1/(2x^2)$ is probably the better $u$-substitution. – eyeballfrog Oct 10 '17 at 3:26

Let t = $1/2x^{2}$
dt = $(-x^{-3}dx)$
So integration converts to $\int {e^{-t}(-2t)dt}$ Now apply by parts