Question

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}.(x+\sqrt{x})dx$ is

• A. $2e^{\sqrt{x}}(x-\sqrt{x}+1)+c$
• B. $e^{\sqrt{x}}(x-2\sqrt{x}+1)+c$
• C. $e^{\sqrt{x}}(x+\sqrt{x})+c$
• D. $e^{\sqrt{x}}(x+\sqrt{x}+1)+c$

Answer 1

The correct option is A $2e^{\sqrt{x}}(x-\sqrt{x}+1)+c$

$Let\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}.(x+\sqrt{x})dx$
$=2\int e^t(t^2+t)dt$
Using by parts
$=2\left[e^t\right(t^2+t)-\int e^t(2t+1\left)dt\right]$
$=2e^t(t^2+t)-2\left[e^t\right(2t+1)-\int e^t(2\left)dt\right]$
$=2e^t(t^2+t)-2e^t(2t+1)+4e^t+c$
$=2e^t(t^2+t-2t-1+2)+c$
$=2e^t(t^2-t+1)+c$
$=2e^{\sqrt{x}}(x-\sqrt{x}+1)+c$