Question

$∫\sqrt{e^x−1}dx$ 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 $I=∫\sqrt{e^x−1}dx=∫\frac{\sqrt{e^x−1}{e}^x}{1+{(\sqrt{e^x−1})}^2}dx$ Put $\sqrt{e^x−1}=t⇒e^x−1=t^2$
$⇒$ $e^{x}dx=2tdt$
$∴$ $I=2∫\frac{t^{2}dt}{1+t^2}=2∫\left(\frac{1+t^2}{1+t^2}\right)dt$
$−2∫\frac{1}{1+t^2}dt$
$=2[t−{\tan }^{−1}t]+c$
$=2[\sqrt{e^x−1}−{\tan }^{−1}\sqrt{e^x−1}]+c$