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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.

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  • $\begingroup$ Try doing a $u$ substitution with $u=1/x$. $\endgroup$ – Sean Lake Oct 10 '17 at 3:26
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    $\begingroup$ $1/(2x^2)$ is probably the better $u$-substitution. $\endgroup$ – eyeballfrog Oct 10 '17 at 3:26
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Let t = $1/2x^{2}$

dt = $(-x^{-3}dx)$

So integration converts to $\int {e^{-t}(-2t)dt}$ Now apply by parts

and then substitute back t.

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  • $\begingroup$ It took a some time to see where you are going with this but it helped thank you $\endgroup$ – Gobabis Oct 10 '17 at 3:52

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