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

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

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integral of 1/(-x-1)

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

- ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = - x - 1

= - ln ∣ - x - 1 ∣

Apply absolute rule:

∣ - a ∣ = ∣ a ∣

∣ - x - 1 ∣ = ∣ x + 1 ∣

= - ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + C

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