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

Substitute into the integral:

\(\int \frac{3x}{3x+1} \, dx = \int \left( \frac{1}{3} - \frac{1}{3u} \right) du\).

Integrate each term separately:

\(\int \frac{1}{3} \, du = \frac{1}{3} u\).

\(\int -\frac{1}{3u} \, du = -\frac{1}{3} \ln|u|\).

Combine the results:

\(\frac{1}{3} u - \frac{1}{3} \ln|u| + c\).

Substitute back \(u = 3x + 1\):

\(\frac{1}{3}(3x+1) - \frac{1}{3} \ln(3x+1) + c\).