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

The integral of (2x-1)/(x^2-3x+2) is ln|4x^2-12x+8|-2ln|2x-2|+2ln|2x-4|+C

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

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

ln 4 x^{2} - 12 x + 8 - 2 ln ∣ 2 x - 2 ∣ + 2 ln ∣ 2 x - 4 ∣ + C

Solution steps

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

Expand

\frac{2 x - 1}{x ^{2} - 3 x + 2} : \frac{2 x}{x ^{2} - 3 x + 2} - \frac{1}{x ^{2} - 3 x + 2}

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

\int \frac{2 x}{x ^{2} - 3 x + 2} d x = ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣

\int \frac{1}{x ^{2} - 3 x + 2} d x = - ln ∣ 2 x - 2 ∣ + ln ∣ 2 x - 4 ∣

= ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - (- ln ∣ 2 x - 2 ∣ + ln ∣ 2 x - 4 ∣)

Simplify

= ln 4 x^{2} - 12 x + 8 - 2 ln ∣ 2 x - 2 ∣ + 2 ln ∣ 2 x - 4 ∣

Add a constant to the solution

= ln 4 x^{2} - 12 x + 8 - 2 ln ∣ 2 x - 2 ∣ + 2 ln ∣ 2 x - 4 ∣ + C

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What is the integral of (2x-1)/(x^2-3x+2) ?
The integral of (2x-1)/(x^2-3x+2) is ln|4x^2-12x+8|-2ln|2x-2|+2ln|2x-4|+C