What is the integral of (3x-5)/(x^2-4x+3) ?

The integral of (3x-5)/(x^2-4x+3) is ln|x-1|+2ln|x-3|+C

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integral of (3x-5)/(x^2-4x+3)

\int \frac{3 x - 5}{x ^{2} - 4 x + 3} d x

ln ∣ x - 1 ∣ + 2 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{3 x - 5}{x ^{2} - 4 x + 3} d x

Take the partial fraction of

\frac{3 x - 5}{x ^{2} - 4 x + 3} : \frac{1}{x - 1} + \frac{2}{x - 3}

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

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

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

= ln ∣ x - 1 ∣ + 2 ln ∣ x - 3 ∣

Add a constant to the solution

= ln ∣ x - 1 ∣ + 2 ln ∣ x - 3 ∣ + C

What is the integral of (3x-5)/(x^2-4x+3) ?
The integral of (3x-5)/(x^2-4x+3) is ln|x-1|+2ln|x-3|+C