What is the integral of 9/(1-3x) ?

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

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integral of 9/(1-3x)

\int \frac{9}{1 - 3 x} d x

- 3 ln ∣ 1 - 3 x ∣ + C

Solution steps

\int \frac{9}{1 - 3 x} d x

Take the constant out:

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

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

Apply u-substitution

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

= 9 \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

= 9 (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = 1 - 3 x

= 9 (- \frac{1}{3} ln ∣ 1 - 3 x ∣)

Simplify

9 (- \frac{1}{3} ln ∣ 1 - 3 x ∣) : - 3 ln ∣ 1 - 3 x ∣

= - 3 ln ∣ 1 - 3 x ∣

Add a constant to the solution

= - 3 ln ∣ 1 - 3 x ∣ + C

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