What is the integral of 1/(ax+b) ?

The integral of 1/(ax+b) is 1/a ln|ax+b|+C

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integral of 1/(ax+b)

\int \frac{1}{a x + b} d x

\frac{1}{a} ln ∣ a x + b ∣ + C

Solution steps

\int \frac{1}{a x + b} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{a} ln ∣ u ∣

Substitute back

u = a x + b

= \frac{1}{a} ln ∣ a x + b ∣

Add a constant to the solution

= \frac{1}{a} ln ∣ a x + b ∣ + C

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