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

The integral of 1/(x(x+1)(x-2)) is -1/2 ln|x|+1/3 ln|x+1|+1/6 ln|x-2|+C

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

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

- \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ( x + 1 ) ( x - 2 )} : - \frac{1}{2 x} + \frac{1}{3 ( x + 1 )} + \frac{1}{6 ( x - 2 )}

= \int - \frac{1}{2 x} + \frac{1}{3 ( x + 1 )} + \frac{1}{6 ( 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}{2 x} d x + \int \frac{1}{3 ( x + 1 )} d x + \int \frac{1}{6 ( x - 2 )} d x

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

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

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + C

\frac{d y}{d x} = \frac{y}{x} + y^{4}

(x^{2} + 3) \frac{d y}{d x} = x y

\int \frac{2 x ^{3} + 4 x ^{2} - 5}{x + 3} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} + 1})

\frac{d}{d x} (3 x^{4} - 4 x^{3} - 12 x^{2} + 5)

What is the integral of 1/(x(x+1)(x-2)) ?
The integral of 1/(x(x+1)(x-2)) is -1/2 ln|x|+1/3 ln|x+1|+1/6 ln|x-2|+C