What is the integral of 1/(ax^2+x) ?

The integral of 1/(ax^2+x) is -ln|ax+1|+ln(|a||x|)+C

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integral of 1/(ax^2+x)

\int \frac{1}{a x ^{2} + x} d x

- ln ∣ a x + 1 ∣ + ln (∣ a ∣ ∣ x ∣) + C

Solution steps

\int \frac{1}{a x ^{2} + x} d x

Factor

\frac{1}{a x ^{2} + x}

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

Apply the Sum Rule:

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - ln ∣ u ∣ + ln ∣ u - 1 ∣

Substitute back

u = a x + 1

= - ln ∣ a x + 1 ∣ + ln ∣ a x + 1 - 1 ∣

ln ∣ a x + 1 - 1 ∣ = ln (∣ a ∣ ∣ x ∣)

= - ln ∣ a x + 1 ∣ + ln (∣ a ∣ ∣ x ∣)

Add a constant to the solution

= - ln ∣ a x + 1 ∣ + ln (∣ a ∣ ∣ x ∣) + C

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What is the integral of 1/(ax^2+x) ?
The integral of 1/(ax^2+x) is -ln|ax+1|+ln(|a||x|)+C