What is the integral of 2/(3y-4) ?

The integral of 2/(3y-4) is 2/3 ln|3y-4|+C

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integral of 2/(3y-4)

\int \frac{2}{3 y - 4} d y

\frac{2}{3} ln ∣ 3 y - 4 ∣ + C

Solution steps

\int \frac{2}{3 y - 4} d y

Take the constant out:

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

= 2 \cdot \int \frac{1}{3 y - 4} d y

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 y - 4

= 2 \cdot \frac{1}{3} ln ∣ 3 y - 4 ∣

Simplify

2 \cdot \frac{1}{3} ln ∣ 3 y - 4 ∣ : \frac{2}{3} ln ∣ 3 y - 4 ∣

= \frac{2}{3} ln ∣ 3 y - 4 ∣

Add a constant to the solution

= \frac{2}{3} ln ∣ 3 y - 4 ∣ + C

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