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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int_{4}^{\infty} e^{- 7 x} d x

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d er i v a t i v e f (x) = 6 x^{2} tan (x)

\frac{d}{d x} (ln ∣ x - 3 ∣)

\frac{d ^{2}}{d x ^{2}} (x + 5 x)

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