माना n geq 2 के लिए f(x) = frac{x}{(1+x^n)^{1 | vedclass

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(A) दिया गया है $f(x) = \frac{x}{(1+x^n)^{1/n}}$.तब $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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) दिया गया है $f(x) = \frac{x}{(1+x^n)^{1/n}}$.तब $f(f(x)) = \frac{f(x)}{(1+f(x)^n)^{1/n}} = \frac{x/(1+x^n)^{1/n}}{(1 + x^n/(1+x^n))^{1/n}} = \frac{x}{(1+x^n+x^n)^{1/n}} = \frac{x}{(1+2x^n)^{1/n}}$.गणितीय आगमन द्वारा,$g(x) = (f \circ f \circ \ldots \circ f)(x) = \frac{x}{(1+nx^n)^{1/n}}$.अब,हमें $I = \int x^{n-2} g(x) \, dx = \int \frac{x^{n-1}}{(1+nx^n)^{1/n}} \, dx$ का मूल्यांकन करना है।माना $u = 1+nx^n$,तो $du = n^2 x^{n-1} \, dx$,जिसका अर्थ है कि $x^{n-1} \, dx = \frac{du}{n^2}$.इन मानों को समाकलन में प्रतिस्थापित करने पर:$I = \int \frac{1}{u^{1/n}} \cdot \frac{du}{n^2} = \frac{1}{n^2} \int u^{-1/n} \, du$.$I = \frac{1}{n^2} \cdot \frac{u^{1 - 1/n}}{1 - 1/n} + K = \frac{1}{n^2} \cdot \frac{u^{(n-1)/n}}{(n-1)/n} + K$.$I = \frac{1}{n^2} \cdot \frac{n}{n-1} u^{(n-1)/n} + K = \frac{1}{n(n-1)} (1+nx^n)^{1 - 1/n} + K$.

माना $n \geq 2$ के लिए $f(x) = \frac{x}{(1+x^n)^{1/n}}$ और $g(x) = \underbrace{(f \circ f \circ \ldots \circ f)}_{n \text{ बार }}(x)$ है। तो $\int x^{n-2} g(x) \, dx$ का मान ज्ञात कीजिए।

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