JEE Main 2019 — Indefinite Integration Question with Solution

$A(x){\left(\sqrt{1-x^2}\right)}^m+C=\int \frac{\sqrt{1-x^2}}{x^4}dx$ $=\int \frac{\sqrt{\frac{1}{x^2}-1}}{x^3}dx$ Let $\frac{1}{x^2}-1=u^2$ $\Rightarrow -\frac{2}{x^3}=\frac{2udu}{dx}$ $\frac{dx}{x^3}=-udu$ $A(x){\left(\sqrt{1-x^2}\right)}^m+C=\int \left(-u^2\right)du=-\frac{u^3}{3}+C$ $\begin{array}{l} =-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{\frac{3}{2}}+C \\ =-\frac{1}{3} \cdot \frac{1}{x^{3}} \cdot\left(1-x^{2}\right)^{\frac{3}{2}}+C \\ =\frac{-1}{3 x^{3}}\left(\sqrt{1-x^{2}}\right)^{3}+C \end{array}$ Compare both sides, $\begin{array}{l} \Rightarrow A(x)=-\frac{1}{3 x^{3}} \text { and } m=3 \\ \Rightarrow(A(x))^{3}=\frac{-1}{27 x^{9}} \end{array}$