What is the integral of (2x-1)^{-1} ?

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

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integral of (2x-1)^{-1}

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

\frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply u-substitution

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

= \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 (2x-1)^{-1} ?
The integral of (2x-1)^{-1} is 1/2 ln|2x-1|+C