What is the integral of 2/(1+3u) ?

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

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integral of 2/(1+3u)

\int \frac{2}{1 + 3 u} d u

\frac{2}{3} ln ∣ 1 + 3 u ∣ + C

Solution steps

\int \frac{2}{1 + 3 u} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{1 + 3 u} d u

Apply u-substitution

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

= 2 \cdot \int \frac{1}{3 v} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

v = 1 + 3 u

= 2 \cdot \frac{1}{3} ln ∣ 1 + 3 u ∣

Simplify

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

= \frac{2}{3} ln ∣ 1 + 3 u ∣

Add a constant to the solution

= \frac{2}{3} ln ∣ 1 + 3 u ∣ + C

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