What is the integral of 8/(2x-1) ?

The integral of 8/(2x-1) is 4ln|2x-1|+C

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integral of 8/(2x-1)

\int \frac{8}{2 x - 1} d x

4 ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{8}{2 x - 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{2 x - 1} d x

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 1

= 8 \cdot \frac{1}{2} ln ∣ 2 x - 1 ∣

Simplify

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

= 4 ln ∣ 2 x - 1 ∣

Add a constant to the solution

= 4 ln ∣ 2 x - 1 ∣ + C

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