Let f(x)=x1+xn1/n for n≥2 and  g(x)=f∘f…ofn times .Then ∫xn−2g(x)dx equal

We have (f∘f)(x)=f(f(x))=f(x)1+(f(x))n1/n=x1+2xn1/nSimilarly, ( fofof )(x)=x1+3xn1/nContinuing in this way we get g(x)=( fof ⋯ of )(x)⏟n times =x1+nxn1/nThus , I=∫xn−2x1+nxn1/ndxPut 1+nxn=tn to obtain I=1n∫tn−1tdt=1n∫tn−2dt=1n(n−1)tn−1+C=1n(n−1)1+nxn1−1/n+C