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
Official answer: A
Compose. 1+f(x)n=1+xn1+xn=1+2xn1+xn1+f(x)^n=1+\dfrac{x^n}{1+x^n}=\dfrac{1+2x^n}{1+x^n}1+f(x)n=1+1+xnxn=1+xn1+2xn, sof(f(x))=f(x)(1+f(x)n)1/n=x(1+2xn)1/n,f(f(x))=\frac{f(x)}{\left(1+f(x)^n\right)^{1/n}}=\frac{x}{(1+2x^n)^{1/n}},f(f(x))=(1+f(x)n)1/nf(x)=(1+2xn)1/nx,and by induction g(x)=x(1+nxn)1/ng(x)=\dfrac{x}{(1+nx^n)^{1/n}}g(x)=(1+nxn)1/nx.
Integrate. ∫xn−2g(x) dx=∫xn−1(1+nxn)−1/ndx\displaystyle\int x^{n-2}g(x)\,dx=\int x^{n-1}(1+nx^n)^{-1/n}dx∫xn−2g(x)dx=∫xn−1(1+nxn)−1/ndx. Put u=1+nxnu=1+nx^nu=1+nxn, du=n2xn−1dxdu=n^2x^{n-1}dxdu=n2xn−1dx:1n2∫u−1/ndu=1n2⋅u1−1n1−1n=1n(n−1)(1+nxn)1−1n+K.\frac1{n^2}\int u^{-1/n}du=\frac1{n^2}\cdot\frac{u^{1-\frac1n}}{1-\frac1n}=\frac1{n(n-1)}\left(1+nx^n\right)^{1-\frac1n}+K.n21∫u−1/ndu=n21⋅1−n1u1−n1=n(n−1)1(1+nxn)1−n1+K.
(A)