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

The integral of ((2x-1))/((x-1)(x-2)(2x-3)) is ln|x-1|+3ln|x-2|-4ln|2x-3|+C

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

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

ln ∣ x - 1 ∣ + 3 ln ∣ x - 2 ∣ - 4 ln ∣ 2 x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 x - 1}{( x - 1 ) ( x - 2 ) ( 2 x - 3 )} : \frac{1}{x - 1} + \frac{3}{x - 2} - \frac{8}{2 x - 3}

= \int \frac{1}{x - 1} + \frac{3}{x - 2} - \frac{8}{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}{x - 1} d x + \int \frac{3}{x - 2} d x - \int \frac{8}{2 x - 3} d x

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

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

\int \frac{8}{2 x - 3} d x = 4 ln ∣ 2 x - 3 ∣

= ln ∣ x - 1 ∣ + 3 ln ∣ x - 2 ∣ - 4 ln ∣ 2 x - 3 ∣

Add a constant to the solution

= ln ∣ x - 1 ∣ + 3 ln ∣ x - 2 ∣ - 4 ln ∣ 2 x - 3 ∣ + C

\int \frac{ln ( x )}{x} d x

\int \frac{1}{t - 1} d t

\frac{d}{d x} ((4 x - 21) ln ((2 x - 11) (x - 5)))

\frac{2}{x ^{2}}, 2

\int \frac{x ^{2} - 18 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

What is the integral of ((2x-1))/((x-1)(x-2)(2x-3)) ?
The integral of ((2x-1))/((x-1)(x-2)(2x-3)) is ln|x-1|+3ln|x-2|-4ln|2x-3|+C