Question

Integration of √(e^x - 1) dx is =?

Answer 1

Dear student
$$\begin{gathered} \int \sqrt{e^x-1}dx \\ Putu=e^x \\ du=e^{x}dx \\ =\int \frac{\sqrt{u-1}du}{u} \\ Applyv=\sqrt{u-1} \\ \frac{dv}{du}=\frac{1}{2\sqrt{u-1}} \\ So,weget \\ =2\int \frac{v^{2}dv}{v^2+1} \\ Consider,\frac{v^2}{v^2+1}=\frac{v^2-\left(v^2+1\right)}{v^2+1}+1 \\ =\frac{-1}{v^2+1}+1 \\ So,2\int \frac{v^{2}dv}{v^2+1}=2\int \left(\frac{-1}{v^2+1}+1\right)dv \\ =2\left(-{\tan }^{-1}v+v\right)+c \\ =2\left(-{\tan }^{-1}\left(\sqrt{u-1}\right)+\left(\sqrt{u-1}\right)\right)+c \\ =2\left(-{\tan }^{-1}\left(\sqrt{e^x-1}\right)+\left(\sqrt{e^x-1}\right)\right)+c \end{gathered}$$
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