Let $f(x)=\frac{x}{\left(1+x^{n}\right)^{1 / n}}$ for $n \geq 2$ and $g(x)=\underbrace{(f \circ f \circ \cdots \circ f)}_{f \text { occurs } n \text { times }}(x)$. Then $\int x^{n-2} g(x) d x$ equals

(A) $\frac{1}{n(n-1)}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+K$

(B) $\frac{1}{n-1}\left(1+n x^{n}\right)^{1-\frac{1}{n}}+K$

(C) $\frac{1}{n(n+1)}\left(1+n x^{n}\right)^{1+\frac{1}{n}}+K$

(D) $\frac{1}{n+1}\left(1+n x^{n}\right)^{1+\frac{1}{n}}+K$

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Indefinite Integrals

Let

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

for

n \geq 2

and

g (x) = f occurs n times (f \circ f \circ \dots \circ f) (x)

. Then

\int x^{n - 2} g (x) d x

equals

Answer: (A)

First, observe the pattern:

By induction,

g (x) = n times f \circ f \circ \dots \circ f (x) = \frac{x}{( 1 + n x ^{n} ) ^{1/ n}}

Therefore:

I = \int x^{n - 2} g (x) d x = \int \frac{x ^{n - 1}}{( 1 + n x ^{n} ) ^{1/ n}} d x

Let

u = 1 + n x^{n}

, then

d u = n^{2} x^{n - 1} d x

I = \frac{1}{n ^{2}} \int u^{- 1/ n} d u = \frac{1}{n ^{2}} \cdot \frac{u ^{1 - 1/ n}}{1 - 1/ n} + K = \frac{1}{n ( n - 1 )} (1 + n x^{n})^{1 - \frac{1}{n}} + K

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option