Integrate (e^(x^2)) / (x^3). By looking at this I'm assuming there is going to be a u substitution followed by an integration by parts. I'm having trouble picking a good u value that will make the integration by parts simpler.

Question

amistre64 · Answer

i dont thnk e^(x^2) can be integrated so that would have to account for the "u" part i believe

amistre64 · Answer

at some point I believe youll get an integral on the right that matches your original to work with

amistre64 · Answer

$$\int \frac{e^{x^2}}{x^3}dx=-\frac{e^{x^2}}{2x^2}-\int \frac{2x\ e^{x^2}}{-2x^2} dx$$

$$\int \frac{e^{x^2}}{x^3}dx=-\frac{e^{x^2}}{2x^2}+\int \frac{\ e^{x^2}}{x} dx$$

hmmmm

amistre64 · Answer

the wolfram give a Ei function but i got no idea what that represents

amistre64 · Answer

http://www.wolframalpha.com/input/?i=Integrate+%28e%5E%28x%5E2%29%29+%2F+%28x%5E3%29

anonymous · Answer

The Ei function on wolfram alpha has something to do with e^(x^2) not being able to be integrated I believe. It makes sense to make that "u". Then dv would have to be (x^-3). V = the integral of (x^-3) which is what you showed I believe. Thank you for your help.

amistre64 · Answer

youre welcome, good luck :)