Integrating both sides, we get:
\[ v(t) = \int (t + 4) dt = \frac{t^2}{2} + 4t + C \]
Given that \( v(0) = 3 \), we can solve for \( C \):
\[ 3 = \frac{0^2}{2} + 4(0) + C \]
\[ C = 3 \]
Therefore, the velocity function is:
\[ v(t) = \frac{t^2}{2} + 4t + 3 \]
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