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(D) To solve the integral $I = \int \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the numerator inside the square root: $I = \int \sqrt{\frac{(1-x)(1-x

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$\sin^{-1} x + \sqrt{1-x^2} + c$

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(D) To solve the integral $I = \int \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the numerator inside the square root: $I = \int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} \, dx = \int \sqrt{\frac{(1-x)^2}{1-x^2}} \, dx$ $I = \int \frac{1-x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-x^2}} \, dx - \int \frac{x}{\sqrt{1-x^2}} \, dx$ For the first part,$\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$. For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$. $-\int \frac{x}{\sqrt{1-x^2}} \, dx = -\int \frac{-1/2 \, du}{\sqrt{u}} = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} = \sqrt{u} = \sqrt{1-x^2}$. Combining these,we get $I = \sin^{-1} x + \sqrt{1-x^2} + c$.

$\int \sqrt{\frac{1-x}{1+x}} \, dx = $

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