Separate the variables:
[tex]y' = \dfrac{dy}{dx} = \dfrac{2x}{1+2y} \implies (1+2y) \, dy = 2x \, dx[/tex]
Integrate both sides:
[tex]\displaystyle \int(1+2y) \, dy = \int 2x \, dx[/tex\
[tex]y + y^2 = x^2 + C[/tex]
Use the given initial condition to solve for C :
[tex]1 + 1^2 = 1^2 + C \implies C = 1[/tex]
So the particular solution is
[tex]\boxed{y + y^2 = x^2 + 1}[/tex]
which you can also solve explicitly for y as a function of x. By completing the square on the left side, we have
[tex]y + y^2 = x^2 + 1[/tex]
[tex]\dfrac14 + y + y^2 = x^2 + \dfrac54[/tex]
[tex]\left(\dfrac12 + y\right)^2 = x^2 + \dfrac54[/tex]
[tex]\dfrac12 + y = \pm \sqrt{x^2+\dfrac54}[/tex]
Note that y(1) = 1 is positive, so the right side should involve the positive square root:
[tex]\dfrac12 + y = \sqrt{x^2+\dfrac54}[/tex]
[tex]\boxed{y = -\dfrac12 + \sqrt{x^2+\dfrac54} = -\dfrac12 + \dfrac12 \sqrt{4x^2+5}}[/tex]
