What is the integral of (-3)/(x+1) ?

The integral of (-3)/(x+1) is -3ln|x+1|+C

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integral of (-3)/(x+1)

\int \frac{- 3}{x + 1} d x

- 3 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{- 3}{x + 1} d x

Take the constant out:

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

= - 3 \cdot \int \frac{1}{x + 1} d x

Apply u-substitution

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

= - 3 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 3 ln ∣ u ∣

Substitute back

u = x + 1

= - 3 ln ∣ x + 1 ∣

Add a constant to the solution

= - 3 ln ∣ x + 1 ∣ + C

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