Question

$\int_{}^{}\sqrt{e^{x}–1}$ dx

• A. 2 $\left\lbrack \sqrt{e^{x}–1}–\tan^{–1}\sqrt{e^{x}–1} \right\rbrack + c$
• B. $\sqrt{e^{x}–1}$– tan^–1$\sqrt{e^{x}–1}$ + c
• C. 2$\left\lbrack \sqrt{e^{x}–1} + \tan^{–1}\sqrt{e^{x}–1} \right\rbrack + c$
• D. $\sqrt{e^{x}–1}$ + tan^–1$\sqrt{e^{x}–1}$ + c

Answer 1

Answer: (A)

2 $\left\lbrack \sqrt{e^{x}–1}–\tan^{–1}\sqrt{e^{x}–1} \right\rbrack + c$

Answer 2

Put e^x–1 = t² ̃ e^x dx = 2t dt

dx = \ dt

= $\int \frac { 2 \mathrm { t } ^ { 2 } } { \mathrm { t } ^ { 2 } + 1 }$ dt = dt

= dt – 2 dt

= 2 $\int \mathrm { dt } - 2 \tan ^ { - 1 } ( \mathrm { t } ) + \mathrm { c }$ = 2t – 2 tan^–1(t) + c

= 2 $\sqrt { \mathrm { e } ^ { \mathrm { x } } - 1 }$ – 2 tan^–1() + c