Question

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

• A. $(x-4\sqrt{x}+4)e^{\sqrt{x}}+c$
• B. $(2x+4\sqrt{x}+4)e^{\sqrt{x}}+c$
• C. $(2x-4\sqrt{x}+4)e^{\sqrt{x}}+c$
• D. $(2x-4\sqrt{x}-4)e^{\sqrt{x}}-c$

Answer 1

The correct option is C $(2x-4\sqrt{x}+4)e^{\sqrt{x}}+c$

Let $I=\int \sqrt{x}{e}^{\sqrt{x}}dx$
Put $\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$\Rightarrow dx=2tdt$
$\therefore I=\int t\cdot 2t\cdot e^{t}dt$
$=2\int t^2{e}^{t}dt$
$=2[t^2{e}^t-2\int e^{t}tdt]$
$=2[t^2{e}^t-2(e^{t}t-e^t)]+c$
$=2[t^2{e}^t-2e^{t}t+2e^t]+c$
$=2[x{e}^{\sqrt{x}}-2e^{\sqrt{x}}\sqrt{x}+2e^{\sqrt{x}}]+c$
$=e^{\sqrt{x}}[2x-4\sqrt{x}+4]+c$