If $\int \frac{\sqrt{1-x^2}}{x^4} \mathrm{~d} x=\mathrm{A}(x)\left(\sqrt{1-x^2}\right)^{\mathrm{m}}+\mathrm{c}$ for a suitable chosen integer $\mathrm{m}$ and a function $\mathrm{A}(x)$, where $\mathrm{c}$ is a constant of integration, then $(\mathrm{A}(x))^{\mathrm{m}}$ equals
$\frac{1}{9 x^4}$
$\frac{-1}{3 x^3}$
$\frac{-1}{27 x^9}$
$\frac{1}{27 x^6}$
Solution
$\text { Let } \begin{aligned}
\mathrm{I} & =\int \frac{\sqrt{1-x^2}}{x^4} \mathrm{~d} x \\
& =\int \frac{x \sqrt{\frac{1}{x^2}-1}}{x^4} \mathrm{~d} x \\
& =\int \frac{\sqrt{\frac{1}{x^2}-1}}{x^3} \mathrm{~d} x
\end{aligned}$
Let $\frac{1}{x^2}-1=\mathrm{t}$
$\begin{aligned}
\therefore \quad \frac{-2}{x^3} & \mathrm{~d} x=\mathrm{dt} \Rightarrow \frac{1}{x^3} \mathrm{~d} x=\frac{-\mathrm{dt}}{2} \\
\therefore \quad \mathrm{I} & =-\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt} \\
& =\frac{-1}{2} \times \frac{(\mathrm{t})^{\frac{3}{2}}}{\frac{3}{2}}+\mathrm{c} \\
& =\frac{-1}{3} \times\left(\frac{1}{x^2}-1\right)^{\frac{3}{2}}+\mathrm{c} \\
& =\frac{-1}{3} \times \frac{\left(1-x^2\right)^{\frac{3}{2}}}{\left(x^2\right)^{\frac{3}{2}}}+\mathrm{c} \\
& =\frac{-1}{3} \times \frac{\left(\sqrt{1-x^2}\right)^3}{x^3}
\end{aligned}$
Comparing with $\mathrm{A}(x)\left(\sqrt{1-x^2}\right)^{\mathrm{m}}+\mathrm{c}$, we get
$\begin{gathered}
\mathrm{A}(x)=\frac{-1}{3 x^3} \text { and } \mathrm{m}=3 \\
\therefore \quad(\mathrm{A}(x))^{\mathrm{m}}=\left(\frac{-1}{3 x^3}\right)^3=\frac{-1}{27 x^9}
\end{gathered}$