Here's the second one.....the partial fraction decomposition deals with repeated linear factors, so we have
[6x - 3] / [(x +1)^2 (x -1)] = A / (x + 1) + B / (x + 1)^2 + C / (x - 1)
And we have
(6x - 3) = A(x+1)(x-1) + B ( x -1) + C (x+1)^2 expand
(6x - 3) = Ax^2 - A + Bx - B + Cx^2 + 2C(x) + C simplify
(6x - 3) = (A + C)x^2 + (B + 2C)x - (A + B - C) ....and, equating coefficients, we have the folllowing system of equations
A + C = 0 → A =- C
B + 2C = 6
A + B - C = 3 → -2C + B = 3
And, adding the second and third equations, we have, 2B = 9 → B = 9/2
Then, using the second equation, (9/2) + 2C = 6 → 2C = 3/2 → C = 3/4 → A -3/4
So this gives us
∫ [6x-3]/ [(x+1)^2 (x-1)] dx =
∫(-3/4)/(x+1) dx + ∫(9/2)/(x + 1)^2 + ∫(3/4)/(x-1) dx
The first and last integrals become
-(3/4)ln(x + 1) + (3/4)ln(x-1) = (3/4)[ln (x - 1) - ln(x+1) = (3/4)ln [(x-1) / (x+1)] + C1
For the middle integral, let x + 1 = u ...then du = dx and we have
(9/2)∫u^(-2) du = -(9/2)u^(-1) = (-9/2)/(x + 1) + C2
And putting everything together (and letting C1 + C2 = C), we arrive at
(3/4)ln [(x-1) / (x+1)] - (-9/2)/(x + 1) + C
Using partial fractions can be helpful in these situations and usually leads to "ln" integrals which can be more easily evaluated...!!!
