int_0^1 sqrt{frac{2+x}{2-x}} , dx = | vedclass

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(D) Let $I = \int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx$. Rationalizing the integrand: $I = \int_0^1 \frac{2+x}{\sqrt{4-x^2}} \, dx = \int_0^1 \frac{2}{\sq

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$\frac{\pi}{3}+2-\sqrt{3}$

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(D) Let $I = \int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx$. Rationalizing the integrand: $I = \int_0^1 \frac{2+x}{\sqrt{4-x^2}} \, dx = \int_0^1 \frac{2}{\sqrt{4-x^2}} \, dx + \int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx$. For the first part,$\int_0^1 \frac{2}{\sqrt{2^2-x^2}} \, dx = 2 \left[ \sin^{-1} \left( \frac{x}{2} \right) \right]_0^1 = 2 \left( \sin^{-1} \frac{1}{2} - \sin^{-1} 0 \right) = 2 \left( \frac{\pi}{6} - 0 \right) = \frac{\pi}{3}$. For the second part,let $t = 4-x^2$,then $dt = -2x \, dx$,so $x \, dx = -\frac{1}{2} \, dt$. $\int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx = -\frac{1}{2} \int_4^3 \frac{1}{\sqrt{t}} \, dt = \frac{1}{2} \int_3^4 t^{-1/2} \, dt = \frac{1}{2} [2\sqrt{t}]_3^4 = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}$. Adding both parts: $I = \frac{\pi}{3} + 2 - \sqrt{3}$.

$\int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx =$

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