Here she is.
$\displaystyle
\int \frac {1}{x^3}e^\frac{1}{x}
$
Thanks.
Using power series rewrite it as $\displaystyle \int\frac{\sum_{n=0}^{\infty}\frac{\bigg(\frac{1}{ x}\bigg)^n}{n!}}{x^3}dx\Rightarrow\int\frac{\sum_{ n=0}^{\infty}\frac{x^{-n}}{n!}}{x^3}dx$
then rewrite it as $\displaystyle \int\sum_{n=0}^{\infty}\frac{x^{-n-3}}{n!}dx$
integrating we get $\displaystyle \int{\frac{e^{\frac{1}{x}}}{x^3}}dx=\sum_{n=0}^{\i nfty}\frac{x^{-n-2}}{(-n-2)n!}$
or you could just use parts whichever way your teacher wants
Note: New question => New thread.
Make the obvious substitution $\displaystyle u^2 = e^x \Rightarrow 2u \frac{du}{dx} = e^x = u^2 \Rightarrow dx = \frac{2 \, du}{u}$.
Then the integral becomes (after a little bit of algebra): $\displaystyle 2 \int \frac{du}{(u+1)^2 u}$.
Now use partial fractions etc.