The given differential equation is:
\[
(x + 2y^3)\frac{dy}{dx} = 2y
\]
We rearrange the equation to the linear form $\frac{dx}{dy} + P(y)x = Q(y)$:
\[
\frac{dx}{dy} = \frac{x + 2y^3}{2y} = \frac{x}{2y} + \frac{2y^3}{2y} = \frac{1}{2y}x + y^2
\]
\[
\frac{dx}{dy} - \frac{1}{2y}x = y^2
\]
This is a linear differential equation in $x$ with $P(y) = -\frac{1}{2y}$.

The integrating factor (IF) is given by:
\[
IF = e^{\int P(y) dy}
\]
Substitute $P(y)$:
\[
IF = e^{\int \left(-\frac{1}{2y}\right) dy} = e^{-\frac{1}{2} \int \frac{1}{y} dy}
\]
\[
IF = e^{-\frac{1}{2} \ln |y|}
\]
Using the logarithm property $a \ln b = \ln b^a$:
\[
IF = e^{\ln |y|^{-\frac{1}{2}}} = |y|^{-\frac{1}{2}}
\]
Assuming $y > 0$, the integrating factor is:
\[
IF = \frac{1}{\sqrt{y}}
\]