Put x=2sin⁡θ. Then dx=2cos⁡θdθ∫x24−x23dx=∫4sin2⁡θ4−4sin2⁡θ32cos⁡θdθ=∫4sin2⁡θ×2cos⁡θ4cos2⁡θ3dθ=∫4sin2⁡θ×2×cos⁡θ(2cos⁡θ)3dθ=∫4sin2⁡θ4cos2⁡θdθ=∫tan2⁡θdθ=∫sec2⁡θ−1dθ=tan⁡θ−θ+c=x4−x2−Sin−1⁡x2+c=x4−x2−Tan−1⁡x4−x2+c.

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$\int \frac{x^{2}}{{(\sqrt{4 - x^{2}})}^{3}} d x =$

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$\frac{x^{2}}{\sqrt{4 - x^{2}}} - {Sin}^{- 1} (\frac{x}{2}) + c$

$\frac{x}{\sqrt{4 - x^{2}}} - {Tan}^{- 1} (\frac{x}{\sqrt{4 - x^{2}}}) + c$

$\frac{x}{\sqrt{4 - x^{2}}} + {Sin}^{- 1} (\frac{2}{\sqrt{4 - x^{2}}}) + c$

$\sqrt{4 - x^{2}} - t a n^{- 1} (\frac{x}{2}) + c$

answer is B.

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Detailed Solution

Put $x = 2 \sin θ . Then d x = 2 \cos θ d θ$

$\int \frac{x^{2}}{{(\sqrt{4 - x^{2}})}^{3}} d x = \int \frac{4 \sin^{2} θ}{{(\sqrt{4 - 4 \sin^{2} θ})}^{3}} 2 \cos θ d θ = \int \frac{4 \sin^{2} θ \times 2 \cos θ}{{(\sqrt{4 \cos^{2} θ})}^{3}} d θ = \int \frac{4 \sin^{2} θ \times 2 \times \cos θ}{(2 \cos θ)^{3}} d θ$

$\begin{array}{l} = \int \frac{4 \sin^{2} θ}{4 \cos^{2} θ} d θ = \int \tan^{2} θ d θ = \int (\sec^{2} θ - 1) d θ = \tan θ - θ + c = \frac{x}{\sqrt{4 - x^{2}}} - {Sin}^{- 1} (\frac{x}{2}) + c \\ = \frac{x}{\sqrt{4 - x^{2}}} - {Tan}^{- 1} (\frac{x}{\sqrt{4 - x^{2}}}) + c . \end{array}$