$\int \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=2 an ^{-1}(\mathrm{f}(x))+\mathrm{c}$ where $x\gt0$ and c is a constant of integration, then $\mathrm{f}(x)$ is

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$\int \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=2 	an ^{-1}(\mathrm{f}(x))+\mathrm{c}$ where $x\gt0$ and c is a constant of integration, then $\mathrm{f}(x)$ is

Aicharya · Accepted Answer

Let $I=\int \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}$ Let $\sqrt{\mathrm{e}^x-1}=\mathrm{t}$ $\begin{array}{ll} 	herefore & \mathrm{e}^x-1=\mathrm{t}^2 \ 	herefore & \mathrm{e}^x=\mathrm{t}^2+1 \ 	herefore & \mathrm{e}^x \mathrm{~d} x=2 \mathrm{t} \mathrm{dt} \ 	herefore & \mathrm{~d} x=\frac{2 \mathrm{t}}{\mathrm{e}^x} \mathrm{dt}=\frac{2 \mathrm{t}}{\mathrm{t}^2+1} \mathrm{dt} \ \begin{aligned} 	herefore & \mathrm{I} \end{aligned}=\int \frac{1}{\mathrm{t}} 	imes \frac{2 \mathrm{t}}{\mathrm{t}^2+1} \mathrm{dt} \ & =2 \int \frac{1}{\mathrm{t}^2+1} \mathrm{dt} \ & =2 	an ^{-1}(\mathrm{t})+\mathrm{c} \ & =2 	an ^{-1}\left(\sqrt{\mathrm{e}^x-1}ight)+\mathrm{c} \ 	herefore & \mathrm{f}(x)=\sqrt{\mathrm{e}^x-1} \end{array}$ $\ldots[$ from (i)]

$\int \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=2 \tan ^{-1}(\mathrm{f}(x))+\mathrm{c}$ where $x\gt0$ and c…

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