What is the integral of 1/(8-3x) ?

The integral of 1/(8-3x) is -1/3 ln|8-3x|+C

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integral of 1/(8-3x)

\int \frac{1}{8 - 3 x} d x

- \frac{1}{3} ln ∣ 8 - 3 x ∣ + C

Solution steps

\int \frac{1}{8 - 3 x} d x

Apply u-substitution

: \int - \frac{1}{3 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 8 - 3 x

= - \frac{1}{3} ln ∣ 8 - 3 x ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ 8 - 3 x ∣ + C

\int_{0}^{1} - ln (x) d x

\frac{d y}{d x} = - y (x + 1)

x + 4 x + 2

x \to 3 lim \frac{\frac{1}{3} x - \frac{1}{3}}{x - 3}

8 cos (x)