If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C,$ for a suitable…
If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C,$ for a suitable chosen integer $\mathrm{m}$ and a function $\mathrm{A}(\mathrm{x})$, where $\mathrm{C}$ is a constant of integration, then $(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}$ equals :
$\frac{-1}{27 x^{9}}$
$\frac{-1}{3 x^{3}}$
$\frac{1}{27 x^{6}}$
$\frac{1}{9 x^{4}}$
Solution
$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x$
$=\int \frac{\sqrt{\frac{1}{x^{2}}-1}}{x^{3}} d x$
Let $\frac{1}{x^{2}}-1=u^{2}$
$\Rightarrow-\frac{2}{x^{3}}=\frac{2 u d u}{d x}$
$\frac{d x}{x^{3}}=-u d u$
$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int\left(-u^{2}\right) d u=-\frac{u^{3}}{3}+C$
$\begin{array}{l}
=-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{\frac{3}{2}}+C \\
=-\frac{1}{3} \cdot \frac{1}{x^{3}} \cdot\left(1-x^{2}\right)^{\frac{3}{2}}+C \\
=\frac{-1}{3 x^{3}}\left(\sqrt{1-x^{2}}\right)^{3}+C
\end{array}$
Compare both sides,
$\begin{array}{l}
\Rightarrow A(x)=-\frac{1}{3 x^{3}} \text { and } m=3 \\
\Rightarrow(A(x))^{3}=\frac{-1}{27 x^{9}}
\end{array}$