What is the integral of (3x-4)/(-x^2+6x-8) ?

The integral of (3x-4)/(-x^2+6x-8) is -3/2 ln|-x^2+6x-8|+5/2 ln|x-2|-5/2 ln|x-4|+C

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integral of (3x-4)/(-x^2+6x-8)

\int \frac{3 x - 4}{- x ^{2} + 6 x - 8} d x

- \frac{3}{2} ln - x^{2} + 6 x - 8 + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{3 x - 4}{- x ^{2} + 6 x - 8} d x

Expand

\frac{3 x - 4}{- x ^{2} + 6 x - 8} : \frac{3 x}{- x ^{2} + 6 x - 8} - \frac{4}{- x ^{2} + 6 x - 8}

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{3 x}{- x ^{2} + 6 x - 8} d x - \int \frac{4}{- x ^{2} + 6 x - 8} d x

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

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

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

Simplify

3 (- \frac{1}{2} ln - x^{2} + 6 x - 8 + 3 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣)) - 4 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣) : - \frac{3}{2} ln - x^{2} + 6 x - 8 + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x - 4 ∣

= - \frac{3}{2} ln - x^{2} + 6 x - 8 + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{3}{2} ln - x^{2} + 6 x - 8 + \frac{5}{2} ln ∣ x - 2 ∣ - \frac{5}{2} ln ∣ x - 4 ∣ + C

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What is the integral of (3x-4)/(-x^2+6x-8) ?
The integral of (3x-4)/(-x^2+6x-8) is -3/2 ln|-x^2+6x-8|+5/2 ln|x-2|-5/2 ln|x-4|+C