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

Answer: (A)

2\sqrt{e^{x} - 1}$– tan^–1$\sqrt{e^{x} - 1} + c

Answer 2

Let I = $\int_{}^{}\sqrt{e^{x} - 1}$dx = $\int_{}^{}\frac{e^{x}\sqrt{e^{x} - 1}}{1 + (\sqrt{e^{x} - 1})^{2}}$dx

Put $\sqrt{e^{x} - 1}$ = t Ž e^x dx = 2t dt

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