What is the integral of (4/(2x-1)) ?

The integral of (4/(2x-1)) is 2ln|2x-1|+C

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integral of (4/(2x-1))

\int (\frac{4}{2 x - 1}) d x

2 ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{4}{2 x - 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{2 x - 1} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= 4 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣

Simplify

4 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣ : 2 ln ∣ 2 x - 1 ∣

= 2 ln ∣ 2 x - 1 ∣

Add a constant to the solution

= 2 ln ∣ 2 x - 1 ∣ + C

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