I need to calculate the expectation, which is the integral of xf(x). So this extra x cancels with the one in the denominator, and then all of that fraction can be taken out as its not dependent on x. To calculate the other part are you suggesting integration by substitution then?
The overall answer should be equal to sigma*e^mu though.
your $\displaystyle f$ is just the density of a log-normally distributed r.v. $\displaystyle (ln\mathcal{N}(\mu,ln(\sigma^2)))$
and the expectation of a log-normally distributed r.v. with these parameters is given by
$\displaystyle e^{\mu+\frac{1}{2}ln \sigma^2$