<s>[INST] How do i solve the integral of $x*\ln(x-1)$? I've been having some trouble trying to solve this integral. I feel like i'm going in circles with this. What i did first is use integration by parts in this manner.
$$\int x\cdot \ln(x-1)\;dx = \ln(x-1)\cdot \frac{1}{2}x^2- \int \frac{1}{x-1}\cdot \frac{1}{2}x^2 \; dx$$
$$= \ln(x-1)\cdot \frac{1}{2}x^2-\frac{1}{2}\int \frac{1}{x-1}\cdot x^2\; dx=$$
So i get $\int \frac{1}{x-1}\cdot x^2 \; dx$ on which i should use integration by parts again i assume.
$$\int \frac{1}{x-1}\cdot x^2\; dx = \frac{1}{1-x} \cdot \frac{1}{2}x^2-\int \frac{-1}{(x-1)^2} \cdot\frac{1}{2}x^2 \; dx$$
And again.
$$\frac{1}{2}\int \frac{-1}{(x-1)^2}\cdot x^2\; dx=x^2 \cdot \frac{-1}{(x-1)^2} - \int x^2 \cdot \frac{1}{x-1}\;dx.$$
$\int x^2 \cdot \frac{1}{x-1}\;dx$ yields  $x^2 \cdot \ln(x-1) - \int 2x\cdot \ln(x-1)\;dx$
I feel like I'm messing something up, however i can't tell what. I've also tried to  use $f(x)=x \space,  g'(x) = \ln(x-1)$ in the first step. Any suggestion is greatly appreciated.
 [/INST] When you're integrating a rational expression, step one is to make sure it's a proper rational expression.  By long division:
$$\frac{x^2}{x-1} = x^2+1+\frac{1}{x-1},$$
which is much easier to integrate.
</s>