If int frac{sqrt{1-x^4}}{x^7} d x=f(x)left{sq | vedclass

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(B) Let $I = \int \frac{\sqrt{1-x^4}}{x^7} dx$. Substitute $x^2 = u$,so $2x dx = du$,which implies $dx = \frac{du}{2x} = \frac{du}{2\sqrt{u}}$. Then $

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$\frac{-1}{216 x^{18}}$

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(B) Let $I = \int \frac{\sqrt{1-x^4}}{x^7} dx$. Substitute $x^2 = u$,so $2x dx = du$,which implies $dx = \frac{du}{2x} = \frac{du}{2\sqrt{u}}$. Then $I = \int \frac{\sqrt{1-u^2}}{u^{7/2}} \cdot \frac{du}{2\sqrt{u}} = \frac{1}{2} \int \frac{\sqrt{1-u^2}}{u^4} du$. Substitute $u = \sin v$,so $du = \cos v dv$. $I = \frac{1}{2} \int \frac{\cos v \cdot \cos v}{\sin^4 v} dv = \frac{1}{2} \int \cot^2 v \csc^2 v dv$. Let $\cot v = w$,then $-\csc^2 v dv = dw$. $I = \frac{1}{2} \int -w^2 dw = -\frac{1}{2} \cdot \frac{w^3}{3} + C = -\frac{1}{6} \cot^3 v + C$. Since $\cot v = \frac{\cos v}{\sin v} = \frac{\sqrt{1-u^2}}{u} = \frac{\sqrt{1-x^4}}{x^2}$,we have $I = -\frac{1}{6} \left( \frac{\sqrt{1-x^4}}{x^2} \right)^3 + C = -\frac{1}{6x^6} (\sqrt{1-x^4})^3 + C$. Comparing with $f(x) \{\sqrt{1-x^4}\}^n + C$,we get $f(x) = -\frac{1}{6x^6}$ and $n = 3$. Thus,$(f(x))^n = (-\frac{1}{6x^6})^3 = -\frac{1}{216x^{18}}$.

If $\int \frac{\sqrt{1-x^4}}{x^7} d x=f(x)\left\{\sqrt{1-x^4}\right\}^n+C$,then $(f(x))^n$ is equal to

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