Question

$\int e^{\sqrt{x}}dx=f\left(x\right)+C;x>0$. Then $f\left(x\right)$ is

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

Answer 1

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

$\int e^{\sqrt{x}}dx=f\left(x\right)+C;x>0$

Let $x=t^2$, then $dx=2tdt$
$\Rightarrow \int e^{\sqrt{x}}dx=\int e^{t}2tdt$

Using integration by parts,
$\int 2t{e}^{t}dt=2t{e}^t-2\int e^{t}dt$
$=2t{e}^t-2e^t+C=2e^t(t-1)+C$
$=2e^{\sqrt{x}}(\sqrt{x}-1)+C$