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

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

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

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

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

Solution steps

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

Expand

\frac{2 x - 3}{x ^{2} - 3 x + 2} : \frac{2 x}{x ^{2} - 3 x + 2} - \frac{3}{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{3}{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{3}{x ^{2} - 3 x + 2} d x = - 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)

= ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - (- 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣))

Simplify

ln 4 x^{2} - 12 x + 8 - 3 ln ∣ 2 x - 2 ∣ + 3 ln ∣ 2 x - 4 ∣ - (- 6 (\frac{1}{2} ln ∣ 2 x - 2 ∣ - \frac{1}{2} ln ∣ 2 x - 4 ∣)) : ln 4 x^{2} - 12 x + 8

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

Add a constant to the solution

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

\frac{d}{d x} (e^{3 x - 2})

\int cot^{3} (5 x) d x

f (x) = 8 x sin (π x)

4 t^{- \frac{1}{8}} + 4 ln (t) + ln (5)

\int (x^{2} + 3 x + 2) e^{x} d x

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