What is the integral of v/(6v^2-1) ?

The integral of v/(6v^2-1) is 1/12 ln|6v^2-1|+C

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integral of v/(6v^2-1)

\int \frac{v}{6 v ^{2} - 1} d v

\frac{1}{12} ln 6 v^{2} - 1 + C

Solution steps

\int \frac{v}{6 v ^{2} - 1} d v

Apply u-substitution

: \int \frac{1}{12 u} d u

= \int \frac{1}{12 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} ln ∣ u ∣

Substitute back

u = 6 v^{2} - 1

= \frac{1}{12} ln 6 v^{2} - 1

Add a constant to the solution

= \frac{1}{12} ln 6 v^{2} - 1 + C

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What is the integral of v/(6v^2-1) ?
The integral of v/(6v^2-1) is 1/12 ln|6v^2-1|+C