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

We have to evaluate the integral $$\displaystyle\int \frac{\sqrt{1-x^{2}}}{x^{4}}\;dx$$ in such a way that it takes the form $$A(x)\bigl(\sqrt{1-x^{2}}\bigr)^{m}+C,$$ where $$m$$ is an integer, $$A(x)$$ is some algebraic function of $$x$$ and $$C$$ is the constant of integration. After finding $$A(x)$$ and $$m$$ we will raise $$A(x)$$ to the power $$m$$ and compare with the options.

To simplify the radical $$\sqrt{1-x^{2}}$$ we introduce the trigonometric substitution $$x=\sin\theta.$$ For this substitution we know the standard relations $$dx=\cos\theta\,d\theta,\qquad \sqrt{1-x^{2}}=\sqrt{1-\sin^{2}\theta}=\cos\theta.$$

Substituting these in the integrand, we get

$$\int\frac{\sqrt{1-x^{2}}}{x^{4}}\;dx=\int\frac{\cos\theta}{(\sin\theta)^{4}}\;dx.$$

But $$dx=\cos\theta\,d\theta,$$ so substituting $$dx$$ as well we obtain

$$\int\frac{\cos\theta}{\sin^{4}\theta}\;\bigl(\cos\theta\,d\theta\bigr)=\int\frac{\cos^{2}\theta}{\sin^{4}\theta}\;d\theta.$$

Now, using the Pythagorean identity $$\cos^{2}\theta=1-\sin^{2}\theta,$$ we rewrite the integrand:

$$\frac{\cos^{2}\theta}{\sin^{4}\theta}=\frac{1-\sin^{2}\theta}{\sin^{4}\theta} =\frac{1}{\sin^{4}\theta}-\frac{\sin^{2}\theta}{\sin^{4}\theta} =\csc^{4}\theta-\csc^{2}\theta.$$

Hence the integral separates as

$$\int\bigl(\csc^{4}\theta-\csc^{2}\theta\bigr)\,d\theta =\int\csc^{4}\theta\,d\theta-\int\csc^{2}\theta\,d\theta.$$

We now evaluate each part. First recall the standard derivative $$\frac{d}{d\theta}\bigl(\cot\theta\bigr)=-\csc^{2}\theta.$$ We use it to integrate $$\csc^{4}\theta$$ by writing one $$\csc^{2}\theta$$ as $$-d(\cot\theta)$$:

$$\int\csc^{4}\theta\,d\theta =\int\csc^{2}\theta\;\csc^{2}\theta\,d\theta =\int\csc^{2}\theta\;\bigl(-d(\cot\theta)\bigr) =-\int\bigl(1+\cot^{2}\theta\bigr)\,d(\cot\theta),$$

because $$\csc^{2}\theta=1+\cot^{2}\theta.$$ Performing this elementary integral gives

$$-\int\bigl(1+\cot^{2}\theta\bigr)\,d(\cot\theta) =-\Bigl[\cot\theta+\frac{\cot^{3}\theta}{3}\Bigr]+C.$$

The second part is straightforward, because $$\int\csc^{2}\theta\,d\theta=-\cot\theta+C.$$

Putting both results together, the original integral becomes

$$\Bigl[-\cot\theta-\frac{\cot^{3}\theta}{3}\Bigr] -\bigl[-\cot\theta\bigr] +C =-\frac{\cot^{3}\theta}{3}+C.$$

We now return to the $$x$$-variable. From the substitution $$x=\sin\theta$$ we have $$\cot\theta=\frac{\cos\theta}{\sin\theta} =\frac{\sqrt{1-x^{2}}}{x}.$$

So the integral in terms of $$x$$ is

$$-\frac{1}{3}\left(\frac{\sqrt{1-x^{2}}}{x}\right)^{3}+C =\Bigl(-\frac{1}{3x^{3}}\Bigr)\bigl(\sqrt{1-x^{2}}\bigr)^{3}+C.$$

Comparing this with the required form $$A(x)\bigl(\sqrt{1-x^{2}}\bigr)^{m}+C,$$ we clearly have $$m=3,\qquad A(x)=-\frac{1}{3x^{3}}.$$

Finally we compute $$(A(x))^{m}=\left(-\frac{1}{3x^{3}}\right)^{3} =\frac{-1}{27x^{9}}.$$

This matches Option A.

Hence, the correct answer is Option A.