Verify a candidate antiderivative and write the general antiderivative

Differentiate $F(x) = \dfrac{x^3}{3} - x$ term by term: the derivative of $\dfrac{x^3}{3}$ is $x^2$ (power rule), and the derivative of $-x$ is $-1$. So $F'(x) = x^2 - 1$.

Since $F'(x) = x^2 - 1 = f(x)$ for all $x$, $F$ is indeed an antiderivative of $f$. Because any two antiderivatives of the same function differ only by a constant, the most general antiderivative of $f(x) = x^2 - 1$ is $\dfrac{x^3}{3} - x + C$, where $C$ is an arbitrary constant.

The letter is C. In part a, once F was verified to satisfy F prime equals f, the general antiderivative was written as x cubed over three minus x plus C. This C represents the arbitrary constant that accounts for all possible antiderivatives of the same function, since any two antiderivatives differ only by a constant.