Answer
$$F(x) = \sqrt{2}x+C$$
Work Step by Step
Given $$f(x) = \sqrt{2}= \sqrt{2}x^{0}$$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
F(x) &= \sqrt{2}x+C
\end{align*}
To check
\begin{align*}
F'(x) &=\sqrt{2}\\
&=f(x)
\end{align*}