integration help!!!!!

Question

∫(x-1)dx/(x+1)

Answer 1

∫(x-1)dx/(x+1) = ∫(x-1+1-1)dx/(x+1)

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

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

                           = x -(2log(x+1))(1) + c    [ Since ∫(1/x)dx = logx ]

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

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

Therefore ∫(x-1)dx/(x+1) = x - 2log(x+1) + c