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Let f(x)=x1+xn1/n for n2 and  g(x)=

ffofn times .Then xn2g(x)dx equal 

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a
1n(n−1)1+nxn1−1/n+k
b
1n−11+nxn1−1/n+k
c
1n(n−1)1+nxn1+1/n+k
d
1n−11+nxn1+1/n+k

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detailed solution

Correct option is A

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


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