Let f(x) = 1 - frac{1}{x}, g_2(x) = f(f(x)), | vedclass

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(B) Step $1$: Find the pattern of $g_n(x)$.$f(x) = 1 - \frac{1}{x} = \frac{x-1}{x}$.$g_2(x) = f(f(x)) = 1 - \frac{1}{1 - 1/x} = 1 - \frac{x}{x-1} = \f

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$-x$

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(B) Step $1$: Find the pattern of $g_n(x)$.$f(x) = 1 - \frac{1}{x} = \frac{x-1}{x}$.$g_2(x) = f(f(x)) = 1 - \frac{1}{1 - 1/x} = 1 - \frac{x}{x-1} = \frac{x-1-x}{x-1} = \frac{-1}{x-1} = \frac{1}{1-x}$.$g_3(x) = f(g_2(x)) = 1 - \frac{1}{1/(1-x)} = 1 - (1-x) = x$.Step $2$: Determine $g_{2025}(x)$ and $g_{2026}(x)$.Since $g_3(x) = x$, the functions cycle with period $3$.$2025$ is divisible by $3$, so $g_{2025}(x) = g_3(x) = x$.$g_{2026}(x) = f(g_{2025}(x)) = f(x) = 1 - \frac{1}{x}$.Step $3$: Solve the integral equation.$\int x \cdot (1 - \frac{1}{x}) \, dx = \int (x - 1) \, dx = \frac{x^2}{2} - x + c$.$\int g_{2025}(x) \, dx = \int x \, dx = \frac{x^2}{2} + c$.Substituting into the given equation: $\frac{x^2}{2} - x + c = \frac{x^2}{2} + c + h(x)$.Thus, $h(x) = -x$.

Let $f(x) = 1 - \frac{1}{x}$, $g_2(x) = f(f(x))$, $g_3(x) = f(f(f(x)))$ and so on. If $\int x \cdot g_{2026}(x) \, dx = \int g_{2025}(x) \, dx + h(x) + c$, then $h(x) = $

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