What is the integral of 1/(8-x) ?

The integral of 1/(8-x) is -ln|8-x|+C

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integral of 1/(8-x)

\int \frac{1}{8 - x} d x

- ln ∣ 8 - x ∣ + C

Solution steps

\int \frac{1}{8 - x} 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 = 8 - x

= - ln ∣ 8 - x ∣

Add a constant to the solution

= - ln ∣ 8 - x ∣ + C

\frac{d}{d x} (7 x + 8)

z e^{z}

n = 0 \sum \infty (\frac{n}{n + 3})^{n}

x \to 1 lim (4 x - 2)