Evaluate the following integrals:

Let $I=\int e^{\sqrt{x}} d x$

$\sqrt{\mathrm{X}}=\mathrm{t} ; \mathrm{X}=\mathrm{t}^{2}$

$\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$

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

Using integration by parts,

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

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

$=2\left(t e^{t}-e^{t}\right)+c$

$=2 e^{t}(t-1)+c$

Replace the value of $\mathrm{t}$

$=2 e^{\sqrt{x}}(\sqrt{x}-1)+c$