What is the integral of (6x-9)/(x^2-8x+12) ?

The integral of (6x-9)/(x^2-8x+12) is -3/4 ln|x-2|+27/4 ln|x-6|+C

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integral of (6x-9)/(x^2-8x+12)

\int \frac{6 x - 9}{x ^{2} - 8 x + 12} d x

- \frac{3}{4} ln ∣ x - 2 ∣ + \frac{27}{4} ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{6 x - 9}{x ^{2} - 8 x + 12} d x

Take the partial fraction of

\frac{6 x - 9}{x ^{2} - 8 x + 12} : - \frac{3}{4 ( x - 2 )} + \frac{27}{4 ( x - 6 )}

= \int - \frac{3}{4 ( x - 2 )} + \frac{27}{4 ( x - 6 )} 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{3}{4 ( x - 2 )} d x + \int \frac{27}{4 ( x - 6 )} d x

\int \frac{3}{4 ( x - 2 )} d x = \frac{3}{4} ln ∣ x - 2 ∣

\int \frac{27}{4 ( x - 6 )} d x = \frac{27}{4} ln ∣ x - 6 ∣

= - \frac{3}{4} ln ∣ x - 2 ∣ + \frac{27}{4} ln ∣ x - 6 ∣

Add a constant to the solution

= - \frac{3}{4} ln ∣ x - 2 ∣ + \frac{27}{4} ln ∣ x - 6 ∣ + C

\int - \frac{1}{( z + 2 ) \cdot ( z + 1 ) z} d z

\int 12 x^{2} - 6 x + 2 d x

\int e^{- 0}

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

\int 380 - 0.8 q d q

What is the integral of (6x-9)/(x^2-8x+12) ?
The integral of (6x-9)/(x^2-8x+12) is -3/4 ln|x-2|+27/4 ln|x-6|+C