_ ^ e^ x ;dx is equal to (A is an arbitrary constant) - Vidyadip

MCQ

$\int_{}^{} {{e^{\sqrt x }}\;dx} $ is equal to

($A $ is an arbitrary constant)

A
${e^{\sqrt x }} + A$
B
$\frac{1}{2}{e^{\sqrt x }} + A$
✓
$2(\sqrt x - 1){e^{\sqrt x }} + A$
D
$2(\sqrt x + 1){e^{\sqrt x }} + A$

✓

Answer

Correct option: C.

$2(\sqrt x - 1){e^{\sqrt x }} + A$

c
(c) $I = \int_{}^{} {{e^{\sqrt x }}.\,dx} $

Put $\sqrt x = t \Rightarrow \frac{1}{{2\sqrt x }}\,dx = dt \Rightarrow dx = 2t\,dt$

$I = \int_{}^{} {{e^t}.\,2t\,dt} = 2\,[t\,.\,{e^t} - {e^t}] + A = 2\,[\sqrt x \,.\,{e^{\sqrt x }} - {e^{\sqrt x }}] + A$

==> $I = 2(\sqrt x - 1)\,.\,{e^{\sqrt x }} + A$.

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