Consider the ODE $$t\,\frac{dy}{dt}=y+2t^3$$ for positive $t$. It can be rewritten in standard form $$\frac{dy}{dt}=\frac{1}{t}y+2t^2$$ and it is in differential form equal to $$dy-\frac{1}{t}y\,dt=2t^2\,dt$$ The natural logarithm $\ln$ is an antiderivative of $\frac{1}{t}$. This suggests $e^{-\ln(t)}=\frac{1}{t}$ as an integrating factor.

But an intellectual guess of this integrating factor based on division of both sides by $t$ in the differential form would be as good because then we get the differential form $$\frac{1}{t}\,dy-\frac{1}{t^2}y=2t\,dt$$ which can be simplified to $$d\left(\frac{1}{t}y\right)= d(t^2)$$ So: $$\frac{y}{t}=t^2+c$$ for some constant $c$. In other words, the general solution of the given ODE is $$y=c\,t+t^3$$