JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

Let I = _ 0 ^ 2 1 3^x + 3 dx Substitute t = 3^x, which gives dt = 3^x 3 dx dx = dt t 3 . When x = 0, t = 1. When x = 2, t = 9. The integral becomes: I = _ 1 ^ 9 1 t + 3 dt t 3 = 1 3 _ 1 ^ 9 1 t(t + 3) dt Using partial fractions: 1 t(t + 3) = 1 3 ( 1 t - 1 t + 3 ) I = 1 3 3 _ 1 ^ 9 ( 1 t - 1 t + 3 ) dt I = 1 3 3 [ t - (t + 3)]_ 1 ^ 9 I = 1 3 3 [ ( t t + 3 ) ]_ 1 ^ 9 I = 1 3 3 ( ( 9 12 ) - ( 1 4 ) ) I = 1 3 3 ( ( 3 4 ) - ( 1 4 ) ) I = 1 3 3 ( 3/4 1/4 ) = 1 3 3 3 I = 1 3