int frac{dx}{sqrt{e^x-1}} = 2 tan^{-1}(f(x)) | vedclass

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(B) Let $I = \int \frac{dx}{\sqrt{e^x-1}}$.Substitute $\sqrt{e^x-1} = t$.Then $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$.Differentiating both sides

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$\sqrt{e^x-1}$

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(B) Let $I = \int \frac{dx}{\sqrt{e^x-1}}$.Substitute $\sqrt{e^x-1} = t$.Then $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$.Differentiating both sides,we get $e^x dx = 2t dt$.Thus,$dx = \frac{2t}{e^x} dt = \frac{2t}{t^2+1} dt$.Substituting these into the integral:$I = \int \frac{1}{t} \cdot \frac{2t}{t^2+1} dt = 2 \int \frac{1}{t^2+1} dt$.Integrating,we get $I = 2 	an^{-1}(t) + c$.Substituting $t = \sqrt{e^x-1}$ back,we get $I = 2 	an^{-1}(\sqrt{e^x-1}) + c$.Comparing this with the given expression $2 	an^{-1}(f(x)) + c$,we find $f(x) = \sqrt{e^x-1}$.

$\int \frac{dx}{\sqrt{e^x-1}} = 2 \tan^{-1}(f(x)) + c$,where $x > 0$ and $c$ is a constant of integration,then $f(x)$ is

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