For -2 lt x lt 2 , ∫√((2-x)/(2+x)) dx is equal to

The correct answer is Option (3) → \(\displaystyle 2\sin^{-1}\left(\frac{x}{2}\right)+\sqrt{4-x^2}+C\)

Since \(2-x \gt 0\), multiplying the numerator and denominator inside the root by \(2-x\) rationalises it:

    \( \displaystyle \sqrt{\frac{2-x}{2+x}} = \frac{2-x}{\sqrt{(2+x)(2-x)}} = \frac{2-x}{\sqrt{4-x^2}} \)

So the integral splits into two standard forms:

    \( \displaystyle I = 2\int\frac{dx}{\sqrt{4-x^2}}-\int\frac{x}{\sqrt{4-x^2}}\,dx \)

    \( \displaystyle 2\int\frac{dx}{\sqrt{4-x^2}} = 2\sin^{-1}\left(\frac{x}{2}\right) \)

With \(t = 4-x^2\), \(dt = -2x\,dx\):

    \( \displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx = -\sqrt{4-x^2} \)

So

    \( \displaystyle I = 2\sin^{-1}\left(\frac{x}{2}\right)+\sqrt{4-x^2}+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle 2\sin^{-1}\left(\frac{x}{2}\right)-\sqrt{4-x^2}+C\) — Incorrect. Its derivative is \(\frac{2+x}{\sqrt{4-x^2}}\), which equals \(\sqrt{\frac{2+x}{2-x}}\), the reciprocal of the integrand.

• Option (2) \(\displaystyle\sin^{-1}\left(\frac{x}{2}\right)+\sqrt{4-x^2}+C\) — Incorrect. The numerator \(2-x\) gives the factor 2 in front of \(\sin^{-1}\left(\frac{x}{2}\right)\).

• Option (4) \(\displaystyle 2\cos^{-1}\left(\frac{x}{2}\right)+\sqrt{4-x^2}+C\) — Incorrect. Its derivative is \(-\frac{2+x}{\sqrt{4-x^2}}\), which is negative throughout.