What is the integral of-1/(x-4) ?

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

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integral of-1/(x-4)

\int - \frac{1}{x - 4} d x

- ln ∣ x - 4 ∣ + C

Solution steps

\int - \frac{1}{x - 4} d x

Take the constant out:

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

= - \int \frac{1}{x - 4} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = x - 4

= - ln ∣ x - 4 ∣

Add a constant to the solution

= - ln ∣ x - 4 ∣ + C

4 x - 32 x, at x = 4

\int_{- 1}^{1} x^{4} d x

n = 1 \sum \infty n^{4} (1 + n^{5})

\int \frac{x ^{3}}{x ^{8} + 25} d x

\frac{d}{d x} ((1 + x)^{n})