Let \(u = 3x + 1\). Then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

The limits of integration change from \(x = 0\) to \(u = 1\) and from \(x = 1\) to \(u = 4\).

Thus, the integral becomes:

\(\int_{1}^{4} \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{4} u^{0.5} \, du\).

Integrate \(u^{0.5}\):

\(\frac{1}{3} \left[ \frac{u^{1.5}}{1.5} \right]_{1}^{4} = \frac{1}{3} \cdot \frac{2}{3} \left[ u^{1.5} \right]_{1}^{4}\).

Evaluate at the bounds:

\(\frac{2}{9} \left[ 4^{1.5} - 1^{1.5} \right] = \frac{2}{9} \left[ 8 - 1 \right] = \frac{2}{9} \cdot 7 = \frac{14}{9}\).

Thus, the answer is \(\frac{14}{9}\) or 1.56.