To evaluate the definite integral \(\int_{1}^{3}(x^2 - 1)dx\), we can first find the antiderivative of the integrand \(x^2 - 1\). The antiderivative of \(x^2\) is \(\frac{x^3}{3}\) and the antiderivative of \(-1\) is \(-x\). So, the antiderivative of \(x^2 - 1\) is \(\frac{x^3}{3} - x + C\), where C is the constant of integration. Using the Fundamental Theorem of Calculus, we have:

\[\int_{1}^{3}(x^2 - 1)dx = \left(\frac{x^3}{3} - x\right)\Big|_{1}^{3} = \left(\frac{3^3}{3} - 3\right) - \left(\frac{1^3}{3} - 1\right) = (9 - 3) - (\frac{1}{3} - 1) = 6 + \frac{2}{3}\)

So, the value of the definite integral is \(6 + \frac{2}{3}\).