Integration help!!!!!!!!!!!!!!!!!!!!!!!!

Question

∫(x+10)dx/(2x^2+5x-3)

Answer 1

We can solve 2x^2 +5x -3

                     = 2x^2 + 6x - x - 3

                     = 2x(x + 3) -1(x + 3)

                     = (x + 3)(2x - 1)

∴ Therefore ʃ (x+10)dx / (2x^2 +5x -3 ) = ʃ (x+10)dx / (x + 3)(2x - 1)

Let (x+10)/(x + 3)(2x - 1) = A/(x + 3) + B/(2x - 1)

 => (x+10)/(x + 3)(2x - 1) = (A(2x-1) + B(x+3)) / (x + 3)(2x - 1)

 =>  x + 10 = A(2x -1) + B(x + 3)

 If x = -3 then 7 = A(7) => A = -1

 If x = 0.5 then 10.5 = B(3.5) => B = 3

Therefore (x+10)/(x + 3)(2x - 1) = -1/(x + 3) + 3/(2x - 1)

Integrating on both sides we get,

                         ʃ (x+10)dx/(x + 3)(2x - 1) = ʃ-1dx/(x + 3) + ʃ3dx/(2x - 1)

                                                               = - ʃ dx/(x + 3) + 3 ʃ dx/(2x - 1)

                                                               = -log(x + 3) + (3)(log(2x - 1))(2) + c [Since ʃdx/x = logx + c]

                                                               =  6log(2x - 1) - log(x + 3) + c

Therefore    ʃ (x+10)dx/(x + 3)(2x - 1) =  6log(2x - 1) - log(x + 3) + c