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

The integral of 6/(9+6x) is ln|9+6x|+C

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integral of 6/(9+6x)

\int \frac{6}{9 + 6 x} d x

ln ∣ 9 + 6 x ∣ + C

Solution steps

\int \frac{6}{9 + 6 x} d x

Take the constant out:

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

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

Apply u-substitution

= 6 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 9 + 6 x

= 6 \cdot \frac{1}{6} ln ∣ 9 + 6 x ∣

Simplify

6 \cdot \frac{1}{6} ln ∣ 9 + 6 x ∣ : ln ∣ 9 + 6 x ∣

= ln ∣ 9 + 6 x ∣

Add a constant to the solution

= ln ∣ 9 + 6 x ∣ + C