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!

The integration should proceed as follows:
$$\color{violet}{I=\int (x+2)\ln(x-3)dx}$$
Assume $x-3=z$ and then we have that $dx=dz$
The integrand hence becomes $$\color{indigo}{I=\int (z+5)\ln z dz}$$
$$\color{blue}{=\int z\ln z dz+\int 5\ln z dz}$$
Integrating the first one by parts and then the second one as usual,
$$\begin{align} & \color{green}{I=\ln z\int z dz - \int\{\frac{d}{dz}(\ln z)\int z dz\}dz+5\cdot\left[\ln z\int  dz - \int\{\frac{d}{dz}(\ln z)\int dz\}dz\right]}
\\ & \color{brown}{=\frac{z^2}{2}\ln z - \int \frac{1}{z}\cdot \frac{z^2}{2} dz + 5\left[z\ln z -z\right]}
\\ & \color{orange}{=\frac{z^2}{2}\ln z - \frac{z^2}{4} + 5\left[z\ln z -z\right] + c}
\\ & \color{red}{=\frac{(x-3)^2}{2}\ln (x-3) - \frac{(x-3)^2}{4} + 5\left[(x-3)\ln (x-3) -(x-3)\right] + c}\end{align}$$
