Evaluate as follows:

Substitute \(6v^{2}-1=t\)

Differentiate with respect to v,

\(12v=\frac{dt}{dv}\)

\(vdv=\frac{dt}{12}\)

Thus, the integral is calculated as:

\(I=\int \frac{vdv}{6v^{2}-1}\)

\(=\int \frac{dt}{12t}\)

\(=\frac{1}{12}\log|t|+c\)

\(=\frac{1}{12}\log|6v^{2}-1|+c\)

Substitute \(6v^{2}-1=t\)

Differentiate with respect to v,

\(12v=\frac{dt}{dv}\)

\(vdv=\frac{dt}{12}\)

Thus, the integral is calculated as:

\(I=\int \frac{vdv}{6v^{2}-1}\)

\(=\int \frac{dt}{12t}\)

\(=\frac{1}{12}\log|t|+c\)

\(=\frac{1}{12}\log|6v^{2}-1|+c\)