The integral ∫ 0 1 2 ln 1 + 2 x 1 + 4 x 2 d x equals

Step-by-step solution

We have, I = ∫ 0 1 2 ln 1 + 2 x 1 + 4 x 2 d x Let, 2 x = tan θ ⇒ d x = 1 2 s e c 2 θ d θ ⇒ I = ∫ 0 π 4 ln 1 + tan θ s e c 2 θ · 1 2 · s e c 2 θ d θ ∵   1 + tan 2 θ = sec 2 θ = 1 2 ∫ 0 π 4 ln 1 + tan θ d θ = 1 2 ∫ 0 π 4 ln 1 + tan π 4 - θ d θ ∵ ∫ 0 a f x d x = ∫ 0 a f a - x d x = 1 2 ∫ 0 π 4 ln 1 + 1 - tan θ 1 + tan θ d θ = 1 2 &#874