JEE Advanced Mathematics Question (2022) — Solution

Step-by-step solution

Let  I = ∫ 1 2 log 2 x 3 + 1 d x + ∫ 1 log 2 9 2 x - 1 1 3 d x Let ∫ 1 log 2 9 2 x - 1 1 3 d x = I 1 Let 2 x - 1 = t 3 ⇒ 2 x ln 2 d x = 3 t 2 d t ⇒ d x = 3 t 2 t 3 + 1 ln 2 d t i.e.  I 1 = ∫ 1 2 3 t 3 ln 2 t 3 + 1 d t   So,  I = ∫ 1 2 log 2 x 3 + 1 d x + ∫ 1 2 3 t 3 ln 2 t 3 + 1 d t or I = ∫ 1 2 log 2 t 3 + 1 + t · 3 t 2 t 3 + 1 ln 2 d t = t ⋅ log 2 t 3 + 1 1 2 = 2 log 2 9 - 1 log 2 2 = 2 log 2 9 - 1 = 4 log 2 3 - 1 So, I = 5