A(x)(1−x2)m+C=∫x41−x2dx =∫x3x21−1dx Let x21−1=u2 ⇒−x32=dx2udu x3dx=−udu A(x)(1−x2)m+C=∫(−u2)du=−3u3+C $\begin{array}{l}
=-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{\frac{3}{2}}+C \
=-\frac{1}{3} \cdot \frac{1}{x^{3}} \cdot\left(1-x^{2}\right)^{\frac{3}{2}}+C \
=\frac{-1}{3 x^{3}}\left(\sqrt{1-x^{2}}\right)^{3}+C
\end{array}Comparebothsides,\begin{array}{l}
\Rightarrow A(x)=-\frac{1}{3 x^{3}} \text { and } m=3 \
\Rightarrow(A(x))^{3}=\frac{-1}{27 x^{9}}
\end{array}$