# Evaluate the following integral: \int\frac{vdv}{6v^{2}-1}

Applications of integrals
Evaluate the following integral: $$\int\frac{vdv}{6v^{2}-1}$$

2021-05-31
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:
$$I=\int \frac{vdv}{6v^{2}-1}$$
$$=\int \frac{dt}{12t}$$
$$=\frac{1}{12}\log|t|+c$$
$$=\frac{1}{12}\log|6v^{2}-1|+c$$