<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] With $u=\ln(x-3)$, $du = \frac{1}{x-3} dx$, $dv = (x-3) dx$, $v = \left(\frac{x^2}{2}+2x\right)$,
$$ \int (x+2) \ln(x-3) dx = \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{x-3} \left(\frac{x^2}{2}+2x\right) dx \\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{w} \left(\frac{(w+3)^2}{2}+2(w+3)\right) dw \\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{w} \left(\frac{w^2+6w+9}{2}+2w+6)\right) dw \\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \left( \frac{1}{2}w + 5 +\frac{21}{2w} \right)dw \\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \frac{w^2}{4} -5 w - \frac{21}{2} \ln w + C\\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \frac{(x-3)^2}{4} -5 (x-3) - \frac{21}{2} \ln (x-3) + C\\ 
= \left(\frac{x^2}{2}+2x - \frac{21}{2}\right) \ln(x-3) - \frac{(x-3)^2}{4}  -5x + C$$
EDIT:  Here's how you can check if your answer is right.  
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