Question

How to integrate (2x-1)/(x+1)^2 dx?

Answer 1

Let (2x - 1) / (x + 1)^2  = A / (x + 1) + B / (x + 1)^2       -> (1)   where A and B are constants

       (2x - 1) / (x + 1)^2  = A / (x + 1) + B / (x + 1)^2

        2x - 1 = A(x + 1) + B

        2x - 1 = Ax + (B + A)

Taking x terms we get, A = 2

If x = 0 then B + A = -1

              => B = -3

Substituting A, B values in (1) we get,

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

 If we integrate, we get,

∫ (2x + 1) / (x+1)^2 dx = ∫ 2 / (x + 1) - ∫3 / (x + 1)^2 dx

                                 = 2 log (x + 1) + 3 / (x + 1) + C  [ Since ∫1/x dx = logx and ∫1/x^2dx = -1/x ]

Therefore ∫ (2x + 1) / (x+1)^2 dx = 2 log (x + 1) + 3 / (x + 1) + C