Let f(x)=x1+xn1/n for n>2 and g(x)= fofo...of (x) ( n times).Then ∫xn−2g(x)dx equals

fofo⁡(x)=f(x)1+(f(x))n1/n=x1+2xn1/nSimilarly fofof⁡(x)=x1+3xn1/n and likewise gx= fof...of (n times )(x)=x1+nxn1/n ∫xn−2g(x)dx=∫xn−2x1+nxn1/ndx =∫xn-11+nxn1/ndx Put 1+nxn=t⇒n2xn-1dx=dt ∫xn−2g(x)dx=∫1n2dtt1n=1n2t1-1n1-1n+c=1n 1+nxn1-1nn-1+c

Let f(x)=x1+xn1/n for n>2 and g(x)= fofo...of (x) ( n times).Then ∫xn−2g(x)dx equals

fofo⁡(x)=f(x)1+(f(x))n1/n=x1+2xn1/nSimilarly fofof⁡(x)=x1+3xn1/n and likewise gx= fof...of (n times )(x)=x1+nxn1/n ∫xn−2g(x)dx=∫xn−2x1+nxn1/ndx =∫xn-11+nxn1/ndx Put 1+nxn=t⇒n2xn-1dx=dt ∫xn−2g(x)dx=∫1n2dtt1n=1n2t1-1n1-1n+c=1n 1+nxn1-1nn-1+c

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Let $f (x) = \frac{x}{{(1 + x^{n})}^{1 / n}} for n > 2 and g (x) = fofo...of (x) (n times) .$ Then $\int x^{n - 2} g (x) dx$ equals

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$\frac{1}{n (n - 1)} {(1 + {nx}^{n})}^{1 + \frac{1}{n}} + K$

$\frac{1}{(n - 1)} {(1 + {nx}^{n})}^{1 + \frac{1}{n}} + K$

$\frac{1}{(n - 1)} {(1 + {nx}^{n})}^{1 - \frac{1}{n}} + K$

$\frac{1}{n (n - 1)} {(1 + {nx}^{n})}^{1 - \frac{1}{n}} + K$

answer is A.

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Detailed Solution

$fofo (x) = \frac{f (x)}{(1 + (f (x))^{n}) 1 / n} = \frac{x}{{(1 + 2 x^{n})}^{1 / n}}$

Similarly $fofof (x) = \frac{x}{{(1 + 3 x^{n})}^{1 / n}}$ and likewise $g (x) = fof...of (n times) (x) = \frac{x}{{(1 + {nx}^{n})}^{1 / n}}$
$\int x^{n - 2} g (x) dx = \int x^{n - 2} \frac{x}{{(1 + {nx}^{n})}^{1 / n}} dx = \int \frac{x^{n - 1}}{{(1 + {nx}^{n})}^{1 / n}} dx$

$Put (1 + {nx}^{n}) =t \Rightarrow n^{2} x^{n - 1} dx = dt \int x^{n - 2} g (x) dx = \int \frac{1}{n^{2}} \frac{d t}{t^{\frac{1}{n}}} = \frac{1}{n^{2}} \frac{t^{1 - \frac{1}{n}}}{1 - \frac{1}{n}} + c = \frac{1}{n} \frac{{(1 + {nx}^{n})}^{1 - \frac{1}{n}}}{n - 1} + c$