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Let f(x)=x(1+xn)1/nf(x)=\dfrac{x}{\left(1+x^n\right)^{1/n}}f(x)=(1+xn)1/nx for n≥2n\ge2n≥2 and g(x)=(f∘f∘⋯∘f)⏟f occurs n times(x)g(x)=\underbrace{(f\circ f\circ\cdots\circ f)}_{f\text{ occurs }n\text{ times}}(x)g(x)=f occurs n times(f∘f∘⋯∘f)(x). Then ∫xn−2g(x) dx\displaystyle\int x^{n-2}g(x)\,dx∫xn−2g(x)dx equals