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

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

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integral of 1/(1+2u)

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

\frac{1}{2} ln ∣ 1 + 2 u ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 v} d v

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

v = 1 + 2 u

= \frac{1}{2} ln ∣ 1 + 2 u ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 1 + 2 u ∣ + C

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