For the given problem:`(dy)/(dx) =2xsqrt(4x^2+1)` is a first order ordinary differential equation in a form of `(dy)/(dx) = f(x,y)` .

 To evaluate this, we rearrange it in a form of variable separable differential equation: `N(y) dy =M(x) dx` .

Cross-multiply `dx ` to the right side:`dy=2xsqrt(4x^2+1)dx` .

Apply direct integration on both sides: `intdy= int 2xsqrt(4x^2+1)dx` .

For the left side, we apply basic integration property: `int (dy)=y` .

For the right side, we may apply u-substitution by letting: `u = 4x^2+1` then `du=8x dx`  or `(du)/8=x dx` .

The integral becomes:

`int 2xsqrt(4x^2+1)dx=int 2sqrt(u)*(du)/8`

                               `= int (sqrt(u)du)/4`

We may apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .

`int (sqrt(u)du)/4= 1/4int sqrt(u)du`

Apply Law of Exponent: `sqrt(x)= x^(1/2)` and Power Rule for integration : int `x^n= x^(n+1)/(n+1)+C` .

`1/4int sqrt(u)du =(1/4) int u^(1/2)du`

                 `=(1/4)u^(1/2+1)/(1/2+1)+C`

                 `=(1/4)u^(3/2)/((3/2)) +C`

                  `=(1/4)u^(3/2)*(2/3) +C`

                 ` =u^(3/2)/6+C`

Plug-in `u=4x^2+1` on `u^(3/2)/6+C` , we get:

`int 2xsqrt(4x^2+1)dx=(4x^2+1)^(3/2)/6+C`

Combining the results from both sides, we get the general solution of the differential equation as:

`y=(4x^2+1)^(3/2)/6+C`