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

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

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

\int \frac{x - 1}{x ( x - 2 )} d x

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

Solution steps

\int \frac{x - 1}{x ( x - 2 )} d x

Expand

\frac{x - 1}{x ( x - 2 )} : \frac{x}{x ( x - 2 )} - \frac{1}{x ( x - 2 )}

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 ( x - 2 )} d x - \int \frac{1}{x ( x - 2 )} d x

\int \frac{x}{x ( x - 2 )} d x = ln ∣ x - 2 ∣

\int \frac{1}{x ( x - 2 )} d x = - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣

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

Simplify

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

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

Add a constant to the solution

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

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