Answer
$$y=(2x^3+1)^3+3$$
Work Step by Step
$$\frac{dy}{dx}=18x^2(2x^3+1)^2$$
$$dy=18x^2(2x^3+1)^2 dx$$
$u=2x^3+1$
$du=6x^2dx$
$3du=18x^2dx$
$$\int dy=\int 3u^2du$$
$$y=u^3+C$$
$$y=(2x^3+1)^3+C$$
We are given that the graph of y passes through the point $(0,4)$. Now, we must plug in these values in order to find the value of $C$.
$4=(0^3+1)^3+C$
$4=1+C$
$C=3$
$$y=(2x^3+1)^3+3$$
