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

The integral of ((2x-1))/((x-2)(x-3)(x-1)) is -3ln|x-2|+5/2 ln|x-3|+1/2 ln|x-1|+C

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

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

- 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 x - 1}{( x - 2 ) ( x - 3 ) ( x - 1 )} : - \frac{3}{x - 2} + \frac{5}{2 ( x - 3 )} + \frac{1}{2 ( x - 1 )}

= \int - \frac{3}{x - 2} + \frac{5}{2 ( x - 3 )} + \frac{1}{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{3}{x - 2} d x + \int \frac{5}{2 ( x - 3 )} d x + \int \frac{1}{2 ( x - 1 )} d x

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

\int \frac{5}{2 ( x - 3 )} d x = \frac{5}{2} ln ∣ x - 3 ∣

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

= - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

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What is the integral of ((2x-1))/((x-2)(x-3)(x-1)) ?
The integral of ((2x-1))/((x-2)(x-3)(x-1)) is -3ln|x-2|+5/2 ln|x-3|+1/2 ln|x-1|+C