<s>[INST] 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!
 [/INST] $\displaystyle \int (x+2)\ln(x-3)dx=\int x\ln(x-3)dx+\int 2\ln(x-3)dx$
Note that we make use of the formulas $\displaystyle\int \ln u du=u\ln u-u+C$ and $\displaystyle\int u\ln udu=\frac{1}{4}u^2(2\ln u-1)+C$, which can be shown via integration by parts.
For the first integral, let $u=x-3$ and $du=dx$ 
This yields $\displaystyle\int (u+3)\ln udu=\int (u\ln u+3\ln u )du=\frac{1}{4}u^2(2\ln u-1)+3u\ln u-u+C_1=\frac{1}{4}(x-3)^2(2\ln (x-3)-1)+3(x-3)\ln (x-3)-(x-3)+C_1$
For the second integral, let $u=x-3$ and $du=dx$
This yields $\displaystyle2\int \ln u du=2u\ln u-2u+C_2=2(x-3)\ln (x-3)-2(x-3)+C_2.$
Therefore $$\displaystyle \int (x+2)\ln(x-3)dx=\frac{1}{4}(x-3)^2(2\ln (x-3)-1)+3(x-3)\ln (x-3)-(x-3)+2(x-3)\ln (x-3)-2(x-3)+C$$
You can further simplify the antiderivative, but this would be the general strategy for integration.
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