# having trouble with integral

• Dec 7th 2010, 06:41 AM
absvalue
having trouble with integral
$\int{\frac{e^\frac{1}{x}}{x^3}dx}$

I initially tried solving with u substitution, $u = 1/x$ and $du = -1/x^2 dx$.

But after I substitute, I'm not sure how to solve:

$\int{\frac{e^u}{u}du}$
• Dec 7th 2010, 07:03 AM
TheEmptySet
Quote:

Originally Posted by absvalue
$\int{\frac{e^\frac{1}{x}}{x^3}dx}$

I initially tried solving with u substitution, $u = 1/x$ and $du = -1/x^2 dx$.

But after I substitute, I'm not sure how to solve:

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

Lets do this as a sequence of two substitutions first let

$\displaystyle u=\frac{1}{x^2} \implies du=-\frac{2}{x^3}dx$

notice that $\displaystyle \sqrt{u}=\frac{1}{x}$ this gives

$\displaystyle -\frac{1}{2}\int e^{\sqrt{u}}du$

Now let $t=\sqrt{u} \implies dt=\frac{1}{2}\cdot \frac{1}{\sqrt{u}}du=\frac{1}{2}\frac{1}{t}du \iff du=2tdt$

Putting this in gives

$\displaystyle -\frac{1}{2}\int e^{\sqrt{u}}du=-\int t e^{t}dt$

Now just integrate by parts and back substitute