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 -ln|x-2|+5/4 ln|x-3|-1/4 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

- ln ∣ x - 2 ∣ + \frac{5}{4} ln ∣ x - 3 ∣ - \frac{1}{4} 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{1}{x - 2} + \frac{5}{4 ( x - 3 )} - \frac{1}{4 ( x + 1 )}

= \int - \frac{1}{x - 2} + \frac{5}{4 ( x - 3 )} - \frac{1}{4 ( 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{1}{x - 2} d x + \int \frac{5}{4 ( x - 3 )} d x - \int \frac{1}{4 ( x + 1 )} d x

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

\int \frac{5}{4 ( x - 3 )} d x = \frac{5}{4} ln ∣ x - 3 ∣

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

= - ln ∣ x - 2 ∣ + \frac{5}{4} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x - 2 ∣ + \frac{5}{4} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ + C

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 -ln|x-2|+5/4 ln|x-3|-1/4 ln|x+1|+C