TITLE: Evaluate $\int{\frac{4x}{(x^2-1)(x-1)}dx}$
QUESTION [5 upvotes]: So I know I need to use the partial fractions method to solve this integral. However when I split it as:
$$\frac{4x}{(x^2-1)(x-1)} = \frac{Ax + B}{x^2-1} + \frac{C}{x-1}$$ 
I find that I can't solve for the values of A, B, C. The question actually hints that first I need to 'factor the denominator completely'. 
I thought it was already factored, but the solution I have factors it to:
$$\frac{4x}{(x^2-1)(x-1)} = \frac{4x}{(x-1)^2(x+1)}$$
Can someone please explain the thought-process behind this transformation, why it is more 'completely factored', and why my initial set-up is incorrect?
Question and solution originally from:
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4b.pdf
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4bsol.pdf

REPLY [5 votes]: Another way of splitting the integrals. Write $4x = (x+1)^{2} - (x-1)^{2}$ then you have 
\begin{align*}
\int\frac{4x}{(x^{2}-1)\cdot(x-1)} \ dx &= \int \frac{ (x+1)^{2}}{(x^{2}-1) \cdot (x-1)} \ dx - \int \frac{(x-1)^{2}}{(x^{2}-1) \cdot (x-1)} \ dx \\ &=\int \frac{x+1}{(x-1)^{2}} \ dx - \int \frac{1}{x+1} \ dx \\ &=\int \frac{1}{x-1} \ dx  + 2 \int\frac{1}{(x-1)^{2}} \ dx - \int\frac{1}{x+1} \ dx
\end{align*}