Answer
\[\frac{x}{\sqrt{4-x^2}}-\sin^{-1}\left(\frac{x}{2}\right)+C\]
Work Step by Step
Let \[I=\int\frac{x^2}{(4-x^2)^{\frac{3}{2}}}dx\;\;\;\ldots (1)\]
Substitute $\;x=2\sin\theta\;\;\;\ldots (2)$
\[\Rightarrow dx=2\cos\theta d\theta\]
(1) becomes
\[I=\int\frac{4\sin^2\theta}{(4-4\sin^2\theta)^{\frac{3}{2}}}2\cos\theta d\theta\]
Simplifying,
\[I=\int\frac{8\sin^2\theta\cos\theta}{8\cos^3\theta}d\theta\]
\[I=\int\tan^2\theta d \theta\]
\[I=\int (\sec^2\theta-1)d\theta\]
\[I=\tan\theta-\theta+C \;\;\;\ldots (3)\]
Where $C$ is constant of integration
From (2)
\[\sin\theta=\frac{x}{2}\]
\[\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{x^2}{4}}\]
\[\cos\theta=\frac{\sqrt{4-x^2}}{2}\]
\[\Rightarrow \tan\theta=\frac{x}{\sqrt{4-x^2}}\]
(3) becomes
\[I=\frac{x}{\sqrt{4-x^2}}-\sin^{-1}\left(\frac{x}{2}\right)+C\]
Hence,\[I=\frac{x}{\sqrt{4-x^2}}-\sin^{-1}\left(\frac{x}{2}\right)+C\;\;.\]
