int frac{d x}{(x-1) sqrt{x^2-1}} is equal to | vedclass

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(C) Let $I = \int \frac{dx}{(x-1)\sqrt{x^2-1}}$.Substitute $x-1 = \frac{1}{t}$,which implies $x = 1 + \frac{1}{t}$.Then,$dx = -\frac{1}{t^2} dt$.Subst

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$-\sqrt{\frac{x+1}{x-1}}+C$

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(C) Let $I = \int \frac{dx}{(x-1)\sqrt{x^2-1}}$.Substitute $x-1 = \frac{1}{t}$,which implies $x = 1 + \frac{1}{t}$.Then,$dx = -\frac{1}{t^2} dt$.Substituting these into the integral:$I = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \sqrt{(1+\frac{1}{t})^2 - 1}} = -\int \frac{dt}{t \sqrt{1 + \frac{1}{t^2} + \frac{2}{t} - 1}} = -\int \frac{dt}{t \sqrt{\frac{1+2t}{t^2}}}$.Since $t = \frac{1}{x-1}$,for $x > 1$,$t > 0$,so $\sqrt{t^2} = t$.$I = -\int \frac{dt}{t \cdot \frac{\sqrt{1+2t}}{t}} = -\int \frac{dt}{\sqrt{1+2t}}$.$I = -\int (1+2t)^{-1/2} dt = -\frac{(1+2t)^{1/2}}{1/2 \cdot 2} + C = -\sqrt{1+2t} + C$.Substituting $t = \frac{1}{x-1}$ back:$I = -\sqrt{1 + \frac{2}{x-1}} + C = -\sqrt{\frac{x-1+2}{x-1}} + C = -\sqrt{\frac{x+1}{x-1}} + C$.

$\int \frac{d x}{(x-1) \sqrt{x^2-1}}$ is equal to

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