_ ^ x e^ x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\sqrt x {e^{\sqrt x }}\;dx = } $

A
$2\sqrt x - {e^{\sqrt x }} - 4\sqrt x \;{e^{\sqrt x }} + c$
✓
$(2x - 4\sqrt x + 4){e^{\sqrt x }} + c$
C
$(2x + 4\sqrt x + 4){e^{\sqrt x }} + c$
D
$(1 - 4\sqrt x ){e^{\sqrt x }} + c$

✓

Answer

Correct option: B.

$(2x - 4\sqrt x + 4){e^{\sqrt x }} + c$

b
(b) $I = \int_{}^{} {\sqrt x .{e^{\sqrt x }}} dx$. Let $x = {t^2} \Rightarrow dx = 2t\,dt$
$\therefore I = 2\int_{}^{} {{t^2}} \,.\,{e^t}dt$ ==> $I = 2({t^2}.{e^t} - 2t{e^t} + 2{e^t}] + c$
==> $I = \frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} + \frac{{3\pi }}{8}} \right)\,} \right| + c$
i.e., $I = {e^{\sqrt x }}[2x - 4\sqrt x + 4] + c$.

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