If int frac{sqrt{1-x^2}}{x^4} ~d x=A(x)left(s | vedclass

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(A) Given $\int \frac{\sqrt{1-x^2}}{x^4} ~d x=A(x)\left(\sqrt{1-x^2}ight)^m+C$. Let $x=\cos 	heta$,then $d x=-\sin 	heta d 	heta$. Substituting t

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$-\frac{1}{27 x^9}$

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(A) Given $\int \frac{\sqrt{1-x^2}}{x^4} ~d x=A(x)\left(\sqrt{1-x^2}ight)^m+C$. Let $x=\cos 	heta$,then $d x=-\sin 	heta d 	heta$. Substituting these into the integral: $\int \frac{\sqrt{1-\cos ^2 	heta}}{\cos ^4 	heta} \cdot(-\sin 	heta d 	heta) = -\int \frac{\sin 	heta}{\cos ^4 	heta} \cdot \sin 	heta d 	heta = -\int 	an ^2 	heta \sec ^2 	heta d 	heta$. Let $u=	an 	heta$,then $d u=\sec ^2 	heta d 	heta$. The integral becomes $-\int u^2 d u = -\frac{u^3}{3}+C = -\frac{	an ^3 	heta}{3}+C$. Since $	an 	heta = \frac{\sqrt{1-\cos ^2 	heta}}{\cos 	heta} = \frac{\sqrt{1-x^2}}{x}$,we have: $-\frac{1}{3}\left(\frac{\sqrt{1-x^2}}{x}ight)^3+C = -\frac{1}{3 x^3}\left(\sqrt{1-x^2}ight)^3+C$. Comparing this with $A(x)\left(\sqrt{1-x^2}ight)^m+C$,we get $A(x)=-\frac{1}{3 x^3}$ and $m=3$. Therefore,$(A(x))^m = \left(-\frac{1}{3 x^3}ight)^3 = -\frac{1}{27 x^9}$.

If $\int \frac{\sqrt{1-x^2}}{x^4} ~d x=A(x)\left(\sqrt{1-x^2}\right)^m+C$ for a suitable chosen integer $m$ and a function $A(x)$,where $C$ is a constant of integration,then $(A(x))^m$ equals

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