Let f(x) = 1 - 1 x , g₂(x) = f(f(x)), g₃(x) = f(f(f(x))) and so on. If x g₂₀₂₆(x) ,dx = g₂₀₂₅(x) ,dx + h(x) + c , then h(x) = ...

Step-by-step solution

Given f(x) = 1 - 1 x = x-1 x g₂(x) = f(f(x)) = 1 - 1 x-1 x = 1 - x x-1 = -1 x-1 = 1 1-x g₃(x) = f(g₂(x)) = 1 - 1 1 1-x = 1 - (1-x) = x g₄(x) = f(g₃(x)) = f(x) The sequence of functions is periodic with a period of 3 . Since 2025 is a multiple of 3 , g₂₀₂₅(x) = g₃(x) = x . Consequently, g₂₀₂₆(x) = g₁(x) = 1 - 1 x . Substituting these into the given integral equation: x g₂₀₂₆(x) ,dx = g₂₀₂₅(x) ,dx + h(x) + c x (1 - 1 x ) dx = x ,dx + h(x) + c (x - 1) ,dx = x^2 2 + h(x) + c x^2 2 - x = x^2 2 + h(x) + c' Comparing both