JEE Challenger
More from Integrals
MathHardAdv 2007 • Paper 2

Integration of Repeated Functional Composition

Let f(x)=x(1+xn)1/nf(x) = \frac{x}{(1 + x^n)^{1/n}} for n≥2n \ge 2 and g(x)=(f∘f∘⋯∘f⏟f occurs n times)(x)g(x) = (\underbrace{f \circ f \circ \dots \circ f}_{f\text{ occurs } n \text{ times}})(x). Then ∫xn−2g(x) dx\int x^{n-2} g(x)\,dx equals

Options

A

1n(n−1)(1+nxn)1−1n+K\frac{1}{n(n-1)}(1 + n x^n)^{1 - \frac{1}{n}} + K

Correct
B

1n−1(1+nxn)1−1n+K\frac{1}{n-1}(1 + n x^n)^{1 - \frac{1}{n}} + K

C

1n(n+1)(1+nxn)1+1n+K\frac{1}{n(n+1)}(1 + n x^n)^{1 + \frac{1}{n}} + K

D

1n+1(1+nxn)1+1n+K\frac{1}{n+1}(1 + n x^n)^{1 + \frac{1}{n}} + K

Topics & Concepts

IntegralsIndefinite IntegrationComposite FunctionsSubstitution Method

Step-by-Step Solution

  1. Iterating the function f(x)f(x):
    f(f(x))=f(x)(1+(f(x))n)1/n=x(1+xn)1/n(1+xn1+xn)1/n=x(1+2xn)1/nf(f(x)) = \frac{f(x)}{(1 + (f(x))^n)^{1/n}} = \frac{\frac{x}{(1+x^n)^{1/n}}}{\left(1 + \frac{x^n}{1+x^n}\right)^{1/n}} = \frac{x}{(1 + 2x^n)^{1/n}}
    By induction, composing ff with itself nn times yields:
    g(x)=(f∘f∘⋯∘f)(x)=x(1+nxn)1/ng(x) = (f \circ f \circ \dots \circ f)(x) = \frac{x}{(1 + n x^n)^{1/n}}

  2. Evaluating the Integral:
    I=∫xn−2g(x) dx=∫xn−1(1+nxn)1/n dxI = \int x^{n-2} g(x)\,dx = \int \frac{x^{n-1}}{(1 + n x^n)^{1/n}}\,dx

Let u=1+nxnu = 1 + n x^n, then du=n2xn−1 dx  ⟹  xn−1 dx=dun2du = n^2 x^{n-1}\,dx \implies x^{n-1}\,dx = \frac{du}{n^2}:
I=1n2∫u−1/n du=1n2⋅u1−1/n1−1/n+K=1n2⋅nn−1u1−1/n+KI = \frac{1}{n^2} \int u^{-1/n}\,du = \frac{1}{n^2} \cdot \frac{u^{1 - 1/n}}{1 - 1/n} + K = \frac{1}{n^2} \cdot \frac{n}{n-1} u^{1 - 1/n} + K
I=1n(n−1)(1+nxn)1−1n+KI = \frac{1}{n(n-1)}(1 + n x^n)^{1 - \frac{1}{n}} + K

Correct Answer: A

PreviousNext