What is the integral of 1/(3y+2) ?

The integral of 1/(3y+2) is 1/3 ln|3y+2|+C

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integral of 1/(3y+2)

\int \frac{1}{3 y + 2} d y

\frac{1}{3} ln ∣ 3 y + 2 ∣ + C

Solution steps

\int \frac{1}{3 y + 2} d y

Apply u-substitution

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 y + 2

= \frac{1}{3} ln ∣ 3 y + 2 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 y + 2 ∣ + C