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

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

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

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

\frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Apply u-substitution

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= \frac{1}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

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What is the integral of (1/(2x-1)) ?
The integral of (1/(2x-1)) is 1/2 ln|2x-1|+C