The integral is
$$\int \frac{-2x+4}{(x^2+1)(x-1)^2}dx$$ So first I split the integral up into parts with A and Bx+C in the numerator like so,
$$\frac{A}{x^2+1}+\frac{Bx+C}{(x-1)^2}$$ but when I do this the integration is off by a lot. Did I need to split the denominator up more or do something else to the numerator?
The usual way to do that is $$\frac{-2x+4}{(x^2+1)(x-1)^2} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1} + \frac{D}{(x-1)^2}$$
Why your way is impossible!! suppose it is true
$$\frac{-2x+4}{(x^2+1)(x-1)^2} = \frac{A}{x^2+1} + \frac{Bx+C}{(x-1)^2} =\frac{ Bx^3+(A+C)x^2 + (B-2A)x+C+A}{(x^2+1)(x-1)^2} $$
Is it possible to find $A,B$ and $C$ such that $$-2x+4 = Bx^3+(A+C)x^2 + (B-2A)x+C+A$$ For all $x$??
You can see if the above holds then $B=0$ because there is no $x^3$ term in the left side and $A+C = 0$ ( no $x^2$ term in the left) and at the same time $A+C = 4$ the constant equal the constant hence $A+C=0=4$ which is a contradiction!