Mathematics - Indefinite Integration Question with Solution | TestHub

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