How can you integrate the function (5x - 1)/(x^(3)-x)?

The first thing to do, is see if it can be simplified: 5x -1 can't, however x^(3) - x simplifies to x(x^2 - 1) and then to x(x-1)(x+1) by the difference of two squares. So now we have (5x - 1)/(x(x-1)(x+1)) which can be split apart into partial fractions. To do this we set (5x - 1)/(x(x-1)(x+1)) = A/x + B/(x-1) + C/(x+1). Multiplying out the denominator on the right gives the equation 5x - 1 = Ax^2 - A + Bx^2 + Bx + Cx^2 - Cx. By comparing coefficients, we can decide that A+B+C=0, B-C=5 and A=-1. A=-1 is given so B+C=-1 adding this to the second equation gives 2B=4 so B=2 and then C=-3.Thus, we now have -1/x + 2/(x-1) -3/(x+1). The integral of this is -ln|x| + 2ln|x-1| - 3ln|x+1|, or ln|(x-1)^2/x(x+1)^3|.