What is the integral of 6/(3x+1) ?

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

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integral of 6/(3x+1)

\int \frac{6}{3 x + 1} d x

2 ln ∣ 3 x + 1 ∣ + C

Solution steps

\int \frac{6}{3 x + 1} 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}{3 x + 1} d x

Apply u-substitution

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

= 6 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 1

= 6 \cdot \frac{1}{3} ln ∣ 3 x + 1 ∣

Simplify

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

= 2 ln ∣ 3 x + 1 ∣

Add a constant to the solution

= 2 ln ∣ 3 x + 1 ∣ + C

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