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!
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 $\displaystyle du= 2xdx$, $\displaystyle \frac{1}{2}du= xdx$.
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.