Let f(x) = frac{x}{(1 + x^7)^{1/7}} and g(x) | vedclass

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(D) Given $f(x) = \frac{x}{(1 + x^7)^{1/7}}$.Calculating $f(f(x)) = \frac{f(x)}{(1 + f(x)^7)^{1/7}} = \frac{x/(1+x^7)^{1/7}}{(1 + x^7/(1+x^7))^{1/7}}

Vedclass · Accepted Answer

$\frac{1}{42} (1 + 7x^7)^{6/7} + C$

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(D) Given $f(x) = \frac{x}{(1 + x^7)^{1/7}}$.Calculating $f(f(x)) = \frac{f(x)}{(1 + f(x)^7)^{1/7}} = \frac{x/(1+x^7)^{1/7}}{(1 + x^7/(1+x^7))^{1/7}} = \frac{x/(1+x^7)^{1/7}}{(1/(1+x^7))^{1/7}} = \frac{x}{(1+2x^7)^{1/7}}$.By induction,$g(x) = (f \circ f \circ f \circ f \circ f \circ f \circ f)(x) = \frac{x}{(1 + 7x^7)^{1/7}}$.We need to evaluate $I = \int x^5 g(x) dx = \int \frac{x^6}{(1 + 7x^7)^{1/7}} dx$.Let $u = 1 + 7x^7$,then $du = 49x^6 dx$,so $x^6 dx = \frac{du}{49}$.$I = \int \frac{1}{u^{1/7}} \cdot \frac{du}{49} = \frac{1}{49} \int u^{-1/7} du = \frac{1}{49} \cdot \frac{u^{6/7}}{6/7} + C = \frac{1}{49} \cdot \frac{7}{6} u^{6/7} + C = \frac{1}{42} (1 + 7x^7)^{6/7} + C$.

Let $f(x) = \frac{x}{(1 + x^7)^{1/7}}$ and $g(x) = (f \circ f \circ f \circ f \circ f \circ f \circ f)(x)$. Then $\int x^5 g(x) dx$ equals (where $C$ is the constant of integration):

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