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

$\begin{array}{l}Let I=\int \sqrt{e^x-1 }dx\text{}\\ Put e^x-1=t^2 \text{⇒}{e}^{x}dx=2t dt\\ \Rightarrow \text{\hspace{0.17em}}dx=\frac{2t}{1+t^2}dt\text{}\\ I=\int t.\frac{2t}{1+t^2}dt\text{}\\ = \int \frac{2t^2}{1+t^2}dt\text{}\\ =\int \frac{2t^2+2-2}{1+t^2}dt\text{}\\ =2\int \frac{t^2+1}{1+t^2}dt-2\int \frac{1}{1+t^2}dt\text{}\\ =2t-2{\tan }^{-1}t+c\text{}\\ =2\left[\sqrt{e^x-1 }-{\tan }^{-1}\sqrt{e^x-1 }\right]+c\end{array}$