Write the primitive or anti-derivative

Given $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}}}$

Let $I=\int f(x) d x$

$\Rightarrow \mathrm{I}=\int\left(\sqrt{\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}}}\right) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int\left(\mathrm{x}^{\frac{1}{2}}+\frac{1}{\mathrm{x}^{\frac{1}{2}}}\right) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int\left(\mathrm{x}^{\frac{1}{2}}+\mathrm{x}^{-\frac{1}{2}}\right) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \mathrm{x}^{\frac{1}{2}} \mathrm{dx}+\int \mathrm{x}^{-\frac{1}{2}} \mathrm{dx}$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow I=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c$

$\Rightarrow I=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+c$

$\Rightarrow I=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c$

$\therefore I=\frac{2}{3} x \sqrt{x}+2 \sqrt{x}+c$

Thus, the primitive of $f(x)$ is $\frac{2}{3} x \sqrt{x}+2 \sqrt{x}+c$