$\displaystyle \int{\frac{e^\frac{1}{x}}{x^3}dx}$

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:

$\displaystyle \int{\frac{e^u}{u}du}$

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- Dec 7th 2010, 06:41 AMabsvaluehaving trouble with integral
$\displaystyle \int{\frac{e^\frac{1}{x}}{x^3}dx}$

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:

$\displaystyle \int{\frac{e^u}{u}du}$ - Dec 7th 2010, 07:03 AMTheEmptySet
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