int sqrt{frac{2+x}{2-x}} , dx is equal to | vedclass

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(D) Let $I = \int \sqrt{\frac{2+x}{2-x}} \, dx$. Multiplying the numerator and denominator inside the square root by $\sqrt{2+x}$: $I = \int \frac{2+x

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \sqrt{\frac{2+x}{2-x}} \, dx$. Multiplying the numerator and denominator inside the square root by $\sqrt{2+x}$: $I = \int \frac{2+x}{\sqrt{(2-x)(2+x)}} \, dx = \int \frac{2+x}{\sqrt{4-x^2}} \, dx$. Splitting the integral into two parts: $I = 2 \int \frac{1}{\sqrt{2^2-x^2}} \, dx + \int \frac{x}{\sqrt{4-x^2}} \, dx$. For the first part,use the standard integral $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}(\frac{x}{a})$. For the second part,let $u = 4-x^2$,so $du = -2x \, dx$ or $x \, dx = -\frac{1}{2} du$. $I = 2 \sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2} \int u^{-1/2} \, du = 2 \sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2} (2u^{1/2}) + C$. $I = 2 \sin^{-1}\left(\frac{x}{2}\right) - \sqrt{4-x^2} + C$.

$\int \sqrt{\frac{2+x}{2-x}} \, dx$ is equal to

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