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!

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- Oct 1st 2012, 04:33 PMdnftpIntegration 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 PMHallsofIvyRe: 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 PMdnftpRe: 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 PMProve ItRe: Integration of exp(x^2)*(x^2-1)/x^3
- Oct 7th 2012, 08:19 PMdnftpRe: 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.