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AIEEE 2007

JEE Main 2007 Mathematics question on Indefinite Integration

If f(x)=x(1+xn)1/nf(x) = \dfrac{x}{(1+x^n)^{1/n}} for n≥2n \geq 2 and g(x)=(fofo…of)⏟f occurs n times(x)g(x) = \underbrace{(f o f o \ldots o f)}_{f \text{ occurs } n \text{ times}}(x). Then ∫xn−2g(x) dx\int x^{n-2} g(x)\, dx equals
  1. A.
    1n(n−1)(1+nxn)1−1n+c\dfrac{1}{n(n-1)}(1+nx^n)^{1-\frac{1}{n}} + c
  2. B.
    1n−1(1+nxn)1−1n+c\dfrac{1}{n-1}(1+nx^n)^{1-\frac{1}{n}} + c
  3. C.
    1n(n+1)(1+nxn)1+1n+c\dfrac{1}{n(n+1)}(1+nx^n)^{1+\frac{1}{n}} + c
  4. D.
    1n+1(1+nxn)1+1n+c\dfrac{1}{n+1}(1+nx^n)^{1+\frac{1}{n}} + c
Show answer and solution

Correct option: A

Solution

First, find f∘f. f(f(x)) = f(x)/( 1 + f(x)n )^(1/n) = [x/(1+xn)^(1/n)] / (1 + xn/(1+xn))^(1/n) = x/(1+2xn)^(1/n).

By induction, applying f n times gives g(x) = x/(1+nxn)^(1/n).

Now compute the integral: ∫ x^(n-2) · x/(1+nxn)^(1/n) dx = ∫ x^(n-1)/(1+nxn)^(1/n) dx.

Let u = 1+nxn, so du = n²x^(n-1)dx, meaning x^(n-1)dx = du/n².

The integral becomes (1/n²) ∫ u^(-1/n) du = (1/n²) · u^(1-1/n)/(1-1/n) + c = (1/n²) · u^((n-1)/n) · n/(n-1) + c = 1/(n(n-1)) · (1+nxn)^(1-1/n) + c.

This matches option 1.

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