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{1 + x}{x ^{2} + 2 x - 1} d x

\frac{d}{d x} (e^{- x} cos (y))

\frac{d}{d x} ((1 + x^{3})^{5})

\frac{\partial}{\partial y} (e^{y} sin (x) y)

\int \frac{x ^{2} sin ( x ) - 2 x + 3}{x ^{2}} 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