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

The integral of 1/(x(ax+b)) is 1/b (-ln((|ax+b|}{|b|})+ln|\frac{ax+b)/b-1|)+C

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integral of 1/(x(ax+b))

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

\frac{1}{b} (- ln (\frac{∣ a x + b ∣}{∣ b ∣}) + ln \frac{a x + b}{b} - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u - b )} d u

= \int \frac{1}{u ( u - b )} d u

u - b = b (\frac{u}{b} - 1)

= \int \frac{1}{u b ( \frac{u}{b} - 1 )} d u

Take the constant out:

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

= \frac{1}{b} \cdot \int \frac{1}{u ( \frac{u}{b} - 1 )} d u

Apply u-substitution

: \int \frac{1}{v ( v - 1 )} d v

= \frac{1}{b} \cdot \int \frac{1}{v ( v - 1 )} d v

Take the partial fraction of

\frac{1}{v ( v - 1 )} : - \frac{1}{v} + \frac{1}{v - 1}

= \frac{1}{b} \cdot \int - \frac{1}{v} + \frac{1}{v - 1} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{b} (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

= \frac{1}{b} (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

Substitute back

= \frac{1}{b} (- ln \frac{a x + b}{b} + ln \frac{a x + b}{b} - 1)

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{a x + b}{b} = \frac{∣ a x + b ∣}{∣ b ∣}

= \frac{1}{b} (- ln (\frac{∣ a x + b ∣}{∣ b ∣}) + ln \frac{a x + b}{b} - 1)

Add a constant to the solution

= \frac{1}{b} (- ln (\frac{∣ a x + b ∣}{∣ b ∣}) + ln \frac{a x + b}{b} - 1) + C

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What is the integral of 1/(x(ax+b)) ?
The integral of 1/(x(ax+b)) is 1/b (-ln((|ax+b|}{|b|})+ln|\frac{ax+b)/b-1|)+C