What is the integral of 1/((x-1)(x-2)(x-3)) ?

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

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integral of 1/((x-1)(x-2)(x-3))

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

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

Add a constant to the solution

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

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