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

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

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integral of 1/(x(x-a))

\int \frac{1}{x ( x - a )} d x

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

Solution steps

\int \frac{1}{x ( x - a )} d x

x - a = a (\frac{x}{a} - 1)

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

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}{x ( \frac{x}{a} - 1 )} d x

Apply u-substitution

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

= \frac{1}{a} \cdot \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}

= \frac{1}{a} \cdot \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

= \frac{1}{a} (- \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 ∣

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

Substitute back

u = \frac{x}{a}

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

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{x}{a} = \frac{∣ x ∣}{∣ a ∣}

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

Add a constant to the solution

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

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y = \frac{x ^{2} - 1}{x ^{2} + 1}

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