Question

$ \int{\sqrt{{{e}^{x}}-1}}\,dx $ is equal to

• A. $ 2[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}]+c $
• B. $ \sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}+c $
• C. $ \sqrt{{{e}^{x}}-1}+{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}+c $
• D. $ 2[\sqrt{{{e}^{x}}-1}+{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}]+c $
• E. $ 2[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}+1}]+c $

Answer 1

Answer: (A)

Let $ I=\int{\sqrt{{{e}^{x}}-1}}dx=\int{\frac{\sqrt{{{e}^{x}}-1}\,{{e}^{x}}}{1+{{(\sqrt{{{e}^{x}}-1})}^{2}}}}dx $ Put $ \sqrt{{{e}^{x}}-1}=t\Rightarrow {{e}^{x}}-1={{t}^{2}} $
$ \Rightarrow $ $ {{e}^{x}}dx=2t\,dt $
$ \therefore $ $ I=2\int{\frac{{{t}^{2}}dt}{1+{{t}^{2}}}}=2\int{\left( \frac{1+{{t}^{2}}}{1+{{t}^{2}}} \right)}dt $
$ -2\int{\frac{1}{1+{{t}^{2}}}}dt $
$ =2[t-{{\tan }^{-1}}t]+c $
$ =2[\sqrt{{{e}^{x}}-1}-{{\tan }^{-1}}\sqrt{{{e}^{x}}-1}]+c $