e^x - 1 dx = - Vidyadip

MCQ

$\int {\sqrt {{e^x} - 1} } dx = $

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

✓

Answer

Correct option: A.

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

a
(a)$A = \int {\sqrt {{e^x} - 1} } \,dx$
Let ${e^x} - 1 = {t^2}$ ==> ${e^x}dx = 2t\,dt$. Hence

$dx = \frac{{2t}}{{{t^2} + 1}}dt$
$A = \int {t\frac{{2t}}{{{t^2} + 1}}} dt = \int {\frac{{2{t^2}}}{{{t^2} + 1}}dt} $
$ = \int {\frac{{2({t^2} + 1) - 2}}{{{t^2} + 1}}dt} $$ = \int {2dt - \int {\frac{{2dt}}{{{t^2} + 1}}} } $
$ = 2t - 2{\tan ^{ - 1}}t + c = 2\sqrt {{e^x} - 1} - 2{\tan ^{ - 1}}\sqrt {{e^x} - 1} + c$.

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