The value of the integral ∫ - log e 2 log e 2 e x log e e x + 1 + e 2 x d x is equal to

Step-by-step solution

Let I = ∫ - log e 2 log e 2 e x log e e x + 1 + e 2 x d x Let us substitute e x = t ⇒ e x d x = d t Lower limit = e - log e 2 = 1 2 Upper limit = e log e 2 = 2 ⇒ I = ∫ 1 2 2 1 × log e t + 1 + t 2 d t Apply integration by-parts. ∫ u t v t d t = u t ∫ v t d t - ∫ u ' t ∫ v t d t d t Take u t = log e t + 1 + t 2 ,   v t = 1 ⇒ u ' t = 1 t + 1 + t 2 1 + 2 t 2 1 + t 2 = t = t ln t 2 + 1 + x 1 2 2 - ∫ 1 / 2 2 t t 2 + 1 d t = t ln t 2 + 1 + t - t