What is the integral of (4x-2)/(x(x-2)(x+1)) ?

The integral of (4x-2)/(x(x-2)(x+1)) is ln|x|+ln|x-2|-2ln|x+1|+C

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integral of (4x-2)/(x(x-2)(x+1))

\int \frac{4 x - 2}{x ( x - 2 ) ( x + 1 )} d x

ln ∣ x ∣ + ln ∣ x - 2 ∣ - 2 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{4 x - 2}{x ( x - 2 ) ( x + 1 )} d x

Take the partial fraction of

\frac{4 x - 2}{x ( x - 2 ) ( x + 1 )} : \frac{1}{x} + \frac{1}{x - 2} - \frac{2}{x + 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ + ln ∣ x - 2 ∣ - 2 ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ + ln ∣ x - 2 ∣ - 2 ln ∣ x + 1 ∣ + C

What is the integral of (4x-2)/(x(x-2)(x+1)) ?
The integral of (4x-2)/(x(x-2)(x+1)) is ln|x|+ln|x-2|-2ln|x+1|+C