Here she is.

$\displaystyle

\int \frac {1}{x^3}e^\frac{1}{x}

$

Thanks.

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- Apr 16th 2008, 04:04 PMGotovina7Not sure how to do this integral.
Here she is.

$\displaystyle

\int \frac {1}{x^3}e^\frac{1}{x}

$

Thanks. - Apr 16th 2008, 04:30 PMMathstud28
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 - Apr 16th 2008, 04:31 PMmr fantastic
- Apr 16th 2008, 04:59 PMGotovina7
Ooh, I got it now! Thanks guys!

- Apr 16th 2008, 05:44 PMGotovina7
How would I do this one?

$\displaystyle \int \frac{1}{e^x+2 \sqrt{e^x}+1}dx $

Thanks :D - Apr 16th 2008, 05:50 PMMathstud28
- Apr 16th 2008, 05:52 PMmr fantastic
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. - Apr 16th 2008, 05:55 PMGotovina7
Ooh, I see it now. Thanks guys. Sorry about the post.

- Apr 16th 2008, 06:14 PMJhevon
- Apr 16th 2008, 06:15 PMGotovina7