$$\int \frac{x^2}{\sqrt{4-x^2}}dx$$
let
$$x=2\sin \theta, dx=2\cos \theta d\theta$$
and from the triangle we get $\tan \theta = \frac{x}{\sqrt{4-x^2}}$
Substitute
$$\int 2x\tan \theta \cos \theta d\theta=2\int \sin \theta \cdot \frac{\sin \theta}{\cos \theta} \cos \theta d\theta$$
$$=2\int 2\sin^2\theta d\theta=2\int 1-\cos 2\theta d\theta=2\theta - 2\sin \theta \cos \theta+c$$
$$2\arcsin \frac{x}{2}-2\cdot \frac{x}{2} \cdot \frac{\sqrt{4-x^2}}{2}+c$$
$$=2\arcsin \frac{x}{2}-\frac{x\sqrt{4-x^2}}{2}+c$$
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