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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u + 1 ) ( u - 1 )} d u

= \int \frac{1}{u ( u + 1 ) ( u - 1 )} d u

Take the partial fraction of

\frac{1}{u ( u + 1 ) ( u - 1 )} : - \frac{1}{u} + \frac{1}{2 ( u + 1 )} + \frac{1}{2 ( u - 1 )}

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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y^{^{''}} + 16 y = 2 sin (2 t)

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

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