Find an antiderivative $F(x)$ with $F^{\prime}(x)=9 x^{4}+20 x+3$ and $F(0)=3$. \[ F(x)= \]

Step 1 ：To find the antiderivative \(F(x)\) of \(F^{\prime}(x)=9 x^{4}+20 x+3\), we need to integrate \(F^{\prime}(x)\) with respect to \(x\).

Step 2 ：The integral of \(9x^{4}\) is \(\frac{9}{5}x^{5}\), the integral of \(20x\) is \(10x^{2}\), and the integral of \(3\) is \(3x\).

Step 3 ：So, we have: \(F(x) = \int F^{\prime}(x) dx = \int (9x^{4} + 20x + 3) dx = \frac{9}{5}x^{5} + 10x^{2} + 3x + C\), where \(C\) is the constant of integration.

Step 4 ：We can find the value of \(C\) by using the condition \(F(0)=3\).

Step 5 ：Substituting \(x=0\) into the equation, we get: \(3 = F(0) = \frac{9}{5}(0)^{5} + 10(0)^{2} + 3(0) + C = C\). So, \(C=3\).

Step 6 ：\(\boxed{F(x) = \frac{9}{5}x^{5} + 10x^{2} + 3x + 3}\) is the antiderivative.