What is the integral of 1/(3x-2) ?

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

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integral of 1/(3x-2)

\int \frac{1}{3 x - 2} d x

\frac{1}{3} ln ∣ 3 x - 2 ∣ + C

Solution steps

\int \frac{1}{3 x - 2} d x

Apply u-substitution

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

= \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 x - 2

= \frac{1}{3} ln ∣ 3 x - 2 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x - 2 ∣ + C

\frac{d}{d x} (12 cos (3 x))

\frac{d y}{d x}, 4 x^{2} + 6 y^{2} = 24 x

\int (x - 3)^{6} d x

\frac{d}{d x} ((a - b x)^{2})

\frac{d}{d x} (3 f (x) + 2 x) x = 5