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(D) To evaluate $I = \int \sqrt{\frac{a+x}{a-x}} \, dx$,multiply the numerator and denominator inside the square root by $\sqrt{a+x}$.$I = \int \frac{

Vedclass · Accepted Answer

$-a \cos^{-1}(x/a) - \sqrt{a^2-x^2} + c$

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(D) To evaluate $I = \int \sqrt{\frac{a+x}{a-x}} \, dx$,multiply the numerator and denominator inside the square root by $\sqrt{a+x}$.$I = \int \frac{a+x}{\sqrt{a^2-x^2}} \, dx = \int \frac{a}{\sqrt{a^2-x^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx$.For the first integral,we use the standard formula $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}(\frac{x}{a}) + c$ or $-\cos^{-1}(\frac{x}{a}) + c$.For the second integral,let $u = a^2-x^2$,then $du = -2x \, dx$,so $\int \frac{x}{\sqrt{a^2-x^2}} \, dx = -\sqrt{a^2-x^2} + c$.Combining these,we get $I = a \sin^{-1}(\frac{x}{a}) - \sqrt{a^2-x^2} + c$.Since $\sin^{-1}(\frac{x}{a}) = \frac{\pi}{2} - \cos^{-1}(\frac{x}{a})$,the expression becomes $a(\frac{\pi}{2} - \cos^{-1}(\frac{x}{a})) - \sqrt{a^2-x^2} + c = -a \cos^{-1}(\frac{x}{a}) - \sqrt{a^2-x^2} + C$.

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

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