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