Definite Integration
Indefinite Integration
Grade Class 12
Question:
Let f(x) = \frac{x}{(1+nx^n)^{1/n}} for n \ge 2 and g(x) = \underbrace{(f \circ f \circ \dots \circ f)}_{f \text{ occurs } n \text{ times}}(x). Then \int x^{n-2} g(x) dx \text{ equals.}
(A) \frac{1}{n(n-1)}(1+nx^n)^{1-\frac{1}{n}}+K
(B) \frac{1}{n-1}(1+nx^n)^{1-\frac{1}{n}}+K
(C) \frac{1}{n(n+1)}(1+nx^n)^{1+\frac{1}{n}}+K
(D) \frac{1}{n+1}(1+nx^n)^{1+\frac{1}{n}}+K
Step-by-Step Solution
Key Concept: Observe the pattern of f(f(x)), f(f(f(x))), etc., to find a general form for g(x), then perform the integration.
Let f(x) = x(1+nx^n)^{-1/n}. Then f(f(x)) = f(x)(1+n(f(x))^n)^{-1/n} = x(1+nx^n)^{-1/n} * (1+n*x^n(1+nx^n)^{-1})^{-1/n} = x(1+nx^n)^{-1/n} * (1+nx^n-nx^n+nx^n(1+nx^n)^{-1})^{-1/n} = x(1+2nx^n)^{-1/n}. By induction, g(x) = x(1+n^2x^n)^{-1/n}. The integral becomes \int x^{n-2} * x(1+n^2x^n)^{-1/n} dx = \int x^{n-1}(1+n^2x^n)^{-1/n} dx. Let u = 1+n^2x^n, du = n^3x^{n-1}dx. The integral is (1/n^3) \int (u-1)^{-1/n} du. This leads to the result in option (A).
Correct Answer: A