The problem asks to find the antiderivative (or indefinite integral) of the function $f(x) = \sqrt{3}$. Let's solve it.

Solution:

The function $f(x) = \sqrt{3}$ is a constant. The antiderivative of a constant $c$ is given by: $\int c \, dx = c x + C$ where $C$ is the constant of integration.

For $f(x) = \sqrt{3}$, the antiderivative is: $\int \sqrt{3} \, dx = \sqrt{3} x + C$

Final Answer:

$F(x) = \sqrt{3} x + C$

Would you like further explanation or examples on antiderivatives?

Here are 5 related questions to deepen understanding:

1. What is the antiderivative of $f(x) = 5$?
2. How do you compute the antiderivative of $f(x) = 3x^2 + 4x + 1$?
3. Can you explain the difference between definite and indefinite integrals?
4. How would you find the antiderivative of $f(x) = e^x$?
5. What is the general rule for finding the antiderivative of a power function $x^n$?

Tip: Always remember to include the constant of integration $C$ when finding indefinite integrals!