Integrate (x-5)/(x+1)(x-2) using partial fractions

We need to split the initial fraction into two. We know that (x-5)/(x+1)(x-2) is equivalent to A/(x+1) + B/(x-2). Multiplying by (x+1)(x-2) we get x-5 = A(x-2) + B(x+1). Expanding the brackets gives x - 5 = Ax -2A +Bx + B. We collect like terms to form the equation x - 5 = (A+B)x + (-2A + B). We compare like terms to form the two equations x = (A+B)x and -5 = (-2A + B). Now that we have two equations, we need to solve them simultaneously. I will be using the method of substitution. Taking equation 1, we divide by x leaving 1 = A + B and rearrange to make A the subject, leaving A = 1 - B. We substitute this into equation 2 leaving -2(1-B) + B = -5. Expanding the bracket leaves -2 + 2B + B = -5. This reduces to 3B = -3. Dividing by 3 leaves B = -1. Substituting this value of B into equation 1 leaves A = 1 - (-1). A = 2.Now we have our two values of A and B, we can integrate the two fractions 2/(x+1) - 1/(x-2). This integrates to 2 ln | x+1 | - ln | x-2 | + C. Combining log rules, this equation simplifies to ln | (x+1)^2/x-2 | + C.