If int {frac{{sqrt {1 - {x^2}} }}{{{x^4}}}} d | vedclass

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(A) We have the integral $I = \int \frac{\sqrt{1-x^{2}}}{x^{4}} dx.$Rewrite the integrand as $I = \int \frac{x \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} dx = \

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

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(A) We have the integral $I = \int \frac{\sqrt{1-x^{2}}}{x^{4}} dx.$Rewrite the integrand as $I = \int \frac{x \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} dx = \int \frac{1}{x^{3}} \sqrt{\frac{1}{x^{2}}-1} dx.$Let $t = \frac{1}{x^{2}} - 1.$ Then $dt = -\frac{2}{x^{3}} dx,$ which implies $\frac{dx}{x^{3}} = -\frac{1}{2} dt.$Substituting these into the integral,we get $I = -\frac{1}{2} \int \sqrt{t} dt = -\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + C = -\frac{1}{3} t^{3/2} + C.$Substituting back $t = \frac{1-x^2}{x^2},$ we get $I = -\frac{1}{3} \left(\frac{1-x^2}{x^2}\right)^{3/2} + C = -\frac{1}{3} \frac{(1-x^2)^{3/2}}{x^3} + C.$Since $(1-x^2)^{3/2} = (\sqrt{1-x^2})^3,$ we have $I = -\frac{1}{3x^3} (\sqrt{1-x^2})^3 + C.$Comparing this with $A(x) (\sqrt{1-x^2})^m + C,$ we identify $m = 3$ and $A(x) = -\frac{1}{3x^3}.$Thus,$(A(x))^m = \left(-\frac{1}{3x^3}\right)^3 = -\frac{1}{27x^9}.$ Note: The provided options seem to have a sign discrepancy or typo; based on the calculation,the result is $-\frac{1}{27x^9}$.

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

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