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. $2[\sqrt{e^x-1}+{\tan }^{-1}\sqrt{e^x-1}]+c$

Answer 1

The correct option is A $2[\sqrt{e^x-1}-{\tan }^{-1}\sqrt{e^x-1}]+c$

$I=\int \sqrt{e^x-1}dx$
Let $e^x-1=t^2$ or $e^{x}dx=2tdt$ or $dx=\frac{2t}{t^2+1}dt$
$\therefore I=\int t\frac{2t}{t^2+1}dt$
$=\int \frac{2t^2}{t^2+1}dt$
$=2\int \frac{t^2+1-1}{t^2+1}dt$
$=2\int (1-\frac{1}{t^2+1})dt$
$=2\int 1dt-2\int \frac{dt}{t^2+1}$
$=2t-2{\tan }^{-1}t+c$
$=2\sqrt{e^x-1}-2{\tan }^{-1}\sqrt{e^x-1}+c$