Answer
$$y==\frac{1}{6}\left(4 x^{2}+1\right)^{3 / 2}+C$$
Work Step by Step
Given$$\frac{d y}{d x} =2 x \sqrt{4 x^{2}+1} $$
by separating the variables we get
$$\ dy = 2 x \sqrt{4 x^{2}+1} \ \ dx$$
by integrating both sides, we get
$$y =\int 2 x \sqrt{4 x^{2}+1} \ \ d x$$
let $u= \sqrt{4 x^{2}+1} \Rightarrow \ du =\frac{4 x}{\sqrt{4 x^{2}+1}} \ dx \Rightarrow u \ du =4x \ dx $$
\Rightarrow 2 x \ dx=\frac{1}{2} u \ du $,
so, we get
\begin{aligned}
y&= \int \sqrt{4 x^{2}+1}(2 x) d x \\ &=\frac{1}{2} \int u^2 \ du \\
&=\frac{1}{6} u^3+C\\
&=\frac{1}{6}\left(4 x^{2}+1\right)^{3 / 2}+C \end{aligned}
