What is the integral of 8/(9+3x) ?

The integral of 8/(9+3x) is 8/3 ln|9+3x|+C

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integral of 8/(9+3x)

\int \frac{8}{9 + 3 x} d x

\frac{8}{3} ln ∣ 9 + 3 x ∣ + C

Solution steps

\int \frac{8}{9 + 3 x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{9 + 3 x} d x

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 9 + 3 x

= 8 \cdot \frac{1}{3} ln ∣ 9 + 3 x ∣

Simplify

8 \cdot \frac{1}{3} ln ∣ 9 + 3 x ∣ : \frac{8}{3} ln ∣ 9 + 3 x ∣

= \frac{8}{3} ln ∣ 9 + 3 x ∣

Add a constant to the solution

= \frac{8}{3} ln ∣ 9 + 3 x ∣ + C

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