Question

$\int \sqrt{x}{e}^{\sqrt{x}}dx$ 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. $(1-4\sqrt{x})e^{\sqrt{x}}+c$

Answer 1

Answer: (B)

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

$I=\int \sqrt{x}{e}^{\sqrt{x}}dx$

Let $\sqrt{x}=t$

So, $x=t^2$ and $dx=2tdt$

$I=\int 2t^2{e}^{t}dt$

$I=2t^2{e}^t-\int \frac{d\left(2t^2\right)}{dt}{e}^{t}dt$

$\therefore I=2t^2{e}^t-\int 4t{e}^{t}dt$

$\therefore I=2t^2{e}^t-4t{e}^t+4\int e^{t}dt$

$\therefore I=2t^2{e}^t-4t{e}^t+4e^t+c$

$\therefore I=e^t(2t^2-4t+4)+c$

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