Let f(x) = frac{x}{(1+x^n)^{1/n}} for n geq 2 | vedclass

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(A) Given $f(x) = \frac{x}{(1+x^n)^{1/n}}$.Then $f(f(x)) = \frac{f(x)}{(1+f(x)^n)^{1/n}} = \frac{x/(1+x^n)^{1/n}}{(1 + x^n/(1+x^n))^{1/n}} = \frac{x}{

Vedclass · Accepted Answer

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

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(A) Given $f(x) = \frac{x}{(1+x^n)^{1/n}}$.Then $f(f(x)) = \frac{f(x)}{(1+f(x)^n)^{1/n}} = \frac{x/(1+x^n)^{1/n}}{(1 + x^n/(1+x^n))^{1/n}} = \frac{x}{(1+x^n+x^n)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$.By induction,$g(x) = (f \circ f \circ \ldots \circ f)(x) = \frac{x}{(1+nx^n)^{1/n}}$.Now,we need to evaluate $I = \int x^{n-2} g(x) \, dx = \int \frac{x^{n-1}}{(1+nx^n)^{1/n}} \, dx$.Let $u = 1+nx^n$,then $du = n^2 x^{n-1} \, dx$,which implies $x^{n-1} \, dx = \frac{du}{n^2}$.Substituting these into the integral:$I = \int \frac{1}{u^{1/n}} \cdot \frac{du}{n^2} = \frac{1}{n^2} \int u^{-1/n} \, du$.$I = \frac{1}{n^2} \cdot \frac{u^{1 - 1/n}}{1 - 1/n} + K = \frac{1}{n^2} \cdot \frac{u^{(n-1)/n}}{(n-1)/n} + K$.$I = \frac{1}{n^2} \cdot \frac{n}{n-1} u^{(n-1)/n} + K = \frac{1}{n(n-1)} (1+nx^n)^{1 - 1/n} + K$.

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$ for $n \geq 2$ and $g(x) = \underbrace{(f \circ f \circ \ldots \circ f)}_{n \text{ times }}(x)$. Then $\int x^{n-2} g(x) \, dx$ equals

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