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

\( \text {Let } \mathrm{I} =\int \frac{\sqrt{1-x^2}}{x^4} \mathrm{~d} x \)
\( =\int \frac{x \sqrt{\frac{1}{x^2}-1}}{x^4} \mathrm{~d} x \)
\( =\int \frac{\sqrt{\frac{1}{x^2}-1}}{x^3} \mathrm{~d} x\)
Let \(\frac{1}{x^2}-1=\mathrm{t}\)
\(\therefore \frac{-2}{x^3} \mathrm{~d} x=\mathrm{dt} \Rightarrow \frac{1}{x^3} \mathrm{~d} x=\frac{-\mathrm{dt}}{2} \)
\( \therefore \mathrm{I} =-\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt} \)
\( =\frac{-1}{2} \times \frac{(\mathrm{t})^{\frac{3}{2}}}{\frac{3}{2}}+\mathrm{c} \)
\( =\frac{-1}{3} \times\left(\frac{1}{x^2}-1\right)^{\frac{3}{2}}+\mathrm{c} \)
\( =\frac{-1}{3} \times \frac{\left(1-x^2\right)^{\frac{3}{2}}}{\left(x^2\right)^{\frac{3}{2}}}+\mathrm{c} \)
\( =\frac{-1}{3} \times \frac{\left(\sqrt{1-x^2}\right)^3}{x^3}\)
Comparing with \(\mathrm{A}(x)\left(\sqrt{1-x^2}\right)^{\mathrm{m}}+\mathrm{c}\), we get
\(\mathrm{A}(x)=\frac{-1}{3 x^3} \text { and } \mathrm{m}=3 \)
\( \therefore (\mathrm{A}(x))^{\mathrm{m}}=\left(\frac{-1}{3 x^3}\right)^3=\frac{-1}{27 x^9}\)