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

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