int sqrt{frac{1+x}{1-x}} , dx = (where C is | vedclass

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(A) To evaluate the integral $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$,we multiply the numerator and denominator inside the square root by $(1+x)$:$I =

Vedclass · Accepted Answer

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

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(A) To evaluate the integral $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$,we multiply the numerator and denominator inside the square root by $(1+x)$:$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 $x \, dx = -\frac{1}{2} du$:$\int \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} \cdot 2u^{1/2} = -\sqrt{1-x^2}$.Combining these,we get $I = \sin^{-1} x - \sqrt{1-x^2} + C$.

$\int \sqrt{\frac{1+x}{1-x}} \, dx = $ (where $C$ is a constant of integration.)

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