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

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

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integral of 1/(8x-3)

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

\frac{1}{8} ln ∣ 8 x - 3 ∣ + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 8 x - 3

= \frac{1}{8} ln ∣ 8 x - 3 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ 8 x - 3 ∣ + C