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

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

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

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

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

Solution steps

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

Expand

x (a - x) : a x - x^{2}

= \int \frac{1}{a x - x ^{2}} d x

Factor

a x - x^{2} : - (- a x + x^{2})

= \int \frac{1}{- ( - a x + x ^{2} )} d x

Take the constant out:

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

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

Factor

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

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

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

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