If ∫ 1 x 1 - x 1 + x d x = g x + c , g 1 = 0 , then g 1 2 is equal to

Step-by-step solution

Given, ∫ 1 x 1 - x 1 + x d x = g x + c Put x = cos 2 θ d x = - 2 sin 2 θ · d θ = ∫ 1 cos 2 θ tan θ - 4 sin θ · cos θ d θ = ∫ 1 cos 2 θ - 4 sin 2 θ d θ = - 2 ∫ 1 - cos 2 θ cos 2 θ d θ = - 2 2 ln sec 2 θ - tan 2 θ + 2 θ + c = ln sec 2 θ - tan 2 θ + 2 θ + c = ln 1 - sin 2 θ cos 2 θ + cos - 1 x + c = ln 1 − 1 − x 2 x + cos − 1 x + c ⏟ g x ∴