What is the integral of 1/(x^2-6x+8) ?

The integral of 1/(x^2-6x+8) is -1/2 ln|x-2|+1/2 ln|x-4|+C

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integral of 1/(x^2-6x+8)

\int \frac{1}{x ^{2} - 6 x + 8} d x

- \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 6 x + 8} d x

Take the partial fraction of

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

= \int - \frac{1}{2 ( x - 2 )} + \frac{1}{2 ( x - 4 )} 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}{2 ( x - 2 )} d x + \int \frac{1}{2 ( x - 4 )} d x

\int \frac{1}{2 ( x - 2 )} d x = \frac{1}{2} ln ∣ x - 2 ∣

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

= - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 4 ∣ + C

\int cos (x) cos (2 x) d x

ln (x + 5)

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

f (x) = x^{2} + 12

(7 x + 2 y) y^{^{'}} = - 2 x - 7 y

What is the integral of 1/(x^2-6x+8) ?
The integral of 1/(x^2-6x+8) is -1/2 ln|x-2|+1/2 ln|x-4|+C