Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx
I'm not sure where to begin because I don't have any integration rules with the lnx
Using the substitution $\displaystyle u = \ln(x^2+2x)$, we have:
$\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{2}\int\left(\frac{x+1}{x^2+2x}\right)\lef t(\frac{x^2+2x}{x+1}\right)\cdot u \;{du} = \frac{1}{2}\int{u}\;{du} = \frac{1}{4}u^2+k.$
Therefore $\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{4}\ln^2(x^2+2x)+k$.