How do i solve the integral of $x*\ln(x-1)$?

Question

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.

Answer 1

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.

Answer 2

Whenever you have an expression in a functional argument, try to get rid of it. For instance, $$\int x\ln(x-1)dx = \int (u+1)\ln(u)du ={\color{red}{ \int u\ln(u)du}} + {\color{blue}{\int \ln(u)du}},\qquad u=x-1$$

Blue is a standard integral ($x\ln(x)-x$). For the red, integrate by parts. It should be trivial afterwards.