How do you integrate $(x+2)\ln(x-3)$? I got $$\left(\frac {x^2}{2} +2x\right)\ln(x-3)-\left(\frac {x^2}{4}-\frac {7x}{2} -\frac {21}{2}\right)\ln(2x-6)$$ as my answer... Not sure If I got it right. Please correct me, thank you!

Hint. You may use an integration by parts
$$
\begin{align}
\int(x+2)\ln(x-3)\:dx&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+2)^2}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+7)(x-3)+25}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\left(x+7+\frac{25}{x-3}\right)\:dx.
\end{align}
$$ Then conclude easily.
