# Integration of exp(x^2)*(x^2-1)/x^3

• Oct 1st 2012, 04:33 PM
dnftp
Integration of exp(x^2)*(x^2-1)/x^3
I am working on a double integral, and I am now stuck on this integral.
exp(x^2)*(x^2-1)/x^3

Wolfram returned this as an answer:
exp(x^2)/2x

Any help or suggestions will be appreciated!
• Oct 1st 2012, 05:17 PM
HallsofIvy
Re: Integration of exp(x^2)*(x^2-1)/x^3
Multiply both numerator and denominator by x to make the integral
$\displaystyle \int \frac{e^{x^2}(x^2-1)}{x^4} xdx$
and use the substitution $\displaystyle u= x^2$ so that $\displaystyle du= 2xdx$, $\displaystyle \frac{1}{2}du= xdx$.
• Oct 5th 2012, 04:17 PM
dnftp
Re: Integration of exp(x^2)*(x^2-1)/x^3
Thanks!
I know the answer is exp(x)/2x, becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?
• Oct 5th 2012, 04:57 PM
Prove It
Re: Integration of exp(x^2)*(x^2-1)/x^3
Quote:

Originally Posted by dnftp
Thanks!
I know the answer is exp(x)/2x, becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?

HallsOfIvy JUST DID! Do the substitution he suggested. If you have any more problems, post your working and we can help you where you are stuck.
• Oct 7th 2012, 08:19 PM
dnftp
Re: Integration of exp(x^2)*(x^2-1)/x^3
I realised that I had wrote the equation wrong...it should be
$\displaystyle \int \frac{e^{x^2}(x^2-1)+1}{x^4} xdx$ using the substitution suggested I got, $\displaystyle \int \frac{e^{u}(u-1)+1}{2u^2} du$= $\displaystyle \frac{e^{u}-1}{2u}$.
Substituting back to x, $\displaystyle \frac{e^{x^2}-1}{2x^2}$. The limits on the integration were from 0 to 2.
$\displaystyle \frac{e^{4}-1}{8} - 0 = 6.6997...$
However, when I do this on wolfram, I get, $\displaystyle \frac{e^{4}-5}{8} = 6.1997...$
I can't see what I am doing wrong.