What is the integral of 1/(9x-1) ?

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

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integral of 1/(9x-1)

\int \frac{1}{9 x - 1} d x

\frac{1}{9} ln ∣ 9 x - 1 ∣ + C

Solution steps

\int \frac{1}{9 x - 1} d x

Apply u-substitution

= \int \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = 9 x - 1

= \frac{1}{9} ln ∣ 9 x - 1 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ 9 x - 1 ∣ + C