What is the integral of-8/(x+4) ?

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

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integral of-8/(x+4)

\int - \frac{8}{x + 4} d x

- 8 ln ∣ x + 4 ∣ + C

Solution steps

\int - \frac{8}{x + 4} 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}{x + 4} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 8 ln ∣ u ∣

Substitute back

u = x + 4

= - 8 ln ∣ x + 4 ∣

Add a constant to the solution

= - 8 ln ∣ x + 4 ∣ + C

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