int sqrt{frac{1+x}{1-x}} , dx = | vedclass

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Question

(C) Let $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$. Multiply the numerator and denominator inside the square root by $(1+x)$: $I = \int \sqrt{\frac{(1+x)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$. Multiply the numerator and denominator inside the square root by $(1+x)$: $I = \int \sqrt{\frac{(1+x)^2}{(1-x)(1+x)}} \, dx = \int \frac{1+x}{\sqrt{1-x^2}} \, dx$. Split the integral into two parts: $I = \int \frac{1}{\sqrt{1-x^2}} \, dx + \int \frac{x}{\sqrt{1-x^2}} \, dx$. The first integral is $\sin^{-1}x$. For the second integral,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$. $\int \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} (2u^{1/2}) = -\sqrt{1-x^2}$. 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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