Question

$\int \sqrt{e^x-1}dx$

Answer 1

$$\begin{gathered} \int \sqrt{e^x-1}dx \\ \mathrm{Let}{e}^x-1=t^2 \\ \Rightarrow e^x=t^2+1 \\ {e}^x=2t\frac{dt}{dx} \\ dx=\frac{2tdt}{e^x} \\ dx=\frac{2tdt}{t^2+1} \\ \mathrm{Now},\int \sqrt{e^x-1}dx \\ =\int \frac{t·2tdt}{t^2+1} \\ =2\int \frac{t^{2}dt}{t^2+1} \\ =2\int \left(\frac{t^2+1-1}{t^2+1}\right)dt \\ =2\int dt-2\int \frac{dt}{t^2+1} \\ =2t-2{\tan }^{-1}\left(t\right)+C \\ =2\sqrt{e^x-1}-2{\tan }^{-1}\left(\sqrt{e^x-1}\right)+C \end{gathered}$$