What is the integral of (x-1)/(x^2-3x) ?

The integral of (x-1)/(x^2-3x) is 1/3 ln|x|+2/3 ln|x-3|+C

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integral of (x-1)/(x^2-3x)

\int \frac{x - 1}{x ^{2} - 3 x} d x

\frac{1}{3} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x - 1}{x ^{2} - 3 x} d x

Expand

\frac{x - 1}{x ^{2} - 3 x} : \frac{x}{x ^{2} - 3 x} - \frac{1}{x ^{2} - 3 x}

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{x}{x ^{2} - 3 x} d x - \int \frac{1}{x ^{2} - 3 x} d x

\int \frac{x}{x ^{2} - 3 x} d x = ln ∣ x - 3 ∣

\int \frac{1}{x ^{2} - 3 x} d x = - \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣

= ln ∣ x - 3 ∣ - (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣)

Simplify

ln ∣ x - 3 ∣ - (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 3 ∣) : \frac{1}{3} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 3 ∣

= \frac{1}{3} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x ∣ + \frac{2}{3} ln ∣ x - 3 ∣ + C

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What is the integral of (x-1)/(x^2-3x) ?
The integral of (x-1)/(x^2-3x) is 1/3 ln|x|+2/3 ln|x-3|+C