$\int \sqrt{\mathrm{e}^x-1} \mathrm{~d} x=$

$\int \sqrt{\mathrm{e}^x-1} \mathrm{~d} x=$
  1. $\sqrt{\mathrm{e}^x-1}+\tan ^{-1} \sqrt{\mathrm{e}^x-1}+\mathrm{c}$, (where c is constant of integration)
  2. $2 \sqrt{\mathrm{e}^x-1}+\tan ^{-1} \sqrt{\mathrm{e}^x-1}+\mathrm{c}$, (where c is constant of integration)
  3. $2 \sqrt{\mathrm{e}^x-1}-2 \tan ^{-1} \sqrt{\mathrm{e}^x-1}+\mathrm{c}$, (where c is constant of integration)
  4. $2 \sqrt{\mathrm{e}^x-1}-\tan ^{-1} \sqrt{\mathrm{e}^x-1}+\mathrm{c}$, (where c is constant of integration)

Solution

$\begin{aligned} & \text { Let } \mathrm{I}=\int \sqrt{\mathrm{e}^x-1} \mathrm{~d} x \\ & \text { Put } \mathrm{e}^x-\mathrm{l}=\mathrm{t}^2 \\ & \Rightarrow \mathrm{e}^x \mathrm{~d} x=2 \mathrm{t} \cdot \mathrm{dt} \\ & \Rightarrow \mathrm{d} x=\frac{2 \mathrm{t}}{\mathrm{t}^2+1} \mathrm{dt} \\ & \therefore \quad \mathrm{I}=\int \mathrm{t} \cdot \frac{2 \mathrm{t}}{\mathrm{t}^2+1} \mathrm{dt}=\int \frac{2 \mathrm{t}^2}{\mathrm{t}^2+1} \mathrm{dt} \\ & \quad=\int \frac{2\left(\mathrm{t}^2+1\right)-2}{\mathrm{t}^2+1} \mathrm{dt} \\ & \quad=2 \int \mathrm{dt}-2 \int \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\ & \quad=2 \mathrm{t}-2 \tan ^{-1} \mathrm{t}+\mathrm{c} \\ & = \\ & \therefore \sqrt{\mathrm{e}^x-1}-2 \tan ^{-1} \sqrt{\mathrm{e}^x-1}+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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