$\displaystyle \int{\frac{e^\frac{1}{x}}{x^3}dx}$
I initially tried solving with u substitution, $\displaystyle u = 1/x$ and $\displaystyle du = -1/x^2 dx$.
But after I substitute, I'm not sure how to solve:
$\displaystyle \int{\frac{e^u}{u}du}$
$\displaystyle \int{\frac{e^\frac{1}{x}}{x^3}dx}$
I initially tried solving with u substitution, $\displaystyle u = 1/x$ and $\displaystyle du = -1/x^2 dx$.
But after I substitute, I'm not sure how to solve:
$\displaystyle \int{\frac{e^u}{u}du}$
Lets do this as a sequence of two substitutions first let
$\displaystyle \displaystyle u=\frac{1}{x^2} \implies du=-\frac{2}{x^3}dx$
notice that $\displaystyle \displaystyle \sqrt{u}=\frac{1}{x}$ this gives
$\displaystyle \displaystyle -\frac{1}{2}\int e^{\sqrt{u}}du$
Now let $\displaystyle t=\sqrt{u} \implies dt=\frac{1}{2}\cdot \frac{1}{\sqrt{u}}du=\frac{1}{2}\frac{1}{t}du \iff du=2tdt$
Putting this in gives
$\displaystyle \displaystyle -\frac{1}{2}\int e^{\sqrt{u}}du=-\int t e^{t}dt$
Now just integrate by parts and back substitute