Q. $\int_{}^{}{\sqrt{x}e^{\sqrt{x}}}dx$ is equal to
A) $2\sqrt{x} - e^{\sqrt{x}} - 4\sqrt{xe^{\sqrt{x}}} + c$
B) $(2x - 4\sqrt{x} + 4)e^{\sqrt{x}} + c$
C) $(2x + 4\sqrt{x} + 4)e^{\sqrt{x}} + c$
D) $(1 - 4\sqrt{x})e^{\sqrt{x}} + c$
Explanation
(b)
Let $I = \int_{}^{}{\sqrt{x}e^{\sqrt{x}}}dx$
Put $x = t^{2}$
$\Rightarrow \ dx = 2t\ dt$
$\therefore\ \ I = 2\int_{}^{}t^{2}e^{t}dt$
$= 2\left\lbrack t^{2}e^{t} - (2t)e^{t} + 2e^{t} \right\rbrack + c$
$= \left( 2x - 4\sqrt{x} + 4 \right)e^{\sqrt{x}} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.