What is the integral of 3/(6x-9) ?

The integral of 3/(6x-9) is 1/2 ln|6x-9|+C

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integral of 3/(6x-9)

\int \frac{3}{6 x - 9} d x

\frac{1}{2} ln ∣ 6 x - 9 ∣ + C

Solution steps

\int \frac{3}{6 x - 9} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{6 x - 9} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 6 x - 9

= 3 \cdot \frac{1}{6} ln ∣ 6 x - 9 ∣

Simplify

3 \cdot \frac{1}{6} ln ∣ 6 x - 9 ∣ : \frac{1}{2} ln ∣ 6 x - 9 ∣

= \frac{1}{2} ln ∣ 6 x - 9 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 6 x - 9 ∣ + C