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

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

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

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

- \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣ + C

f (x) = x^{2} - 20 x, g (x) = 0

\frac{d}{d x} (ln (\frac{1}{6 x}))

(ln (1 - 7 x))^{^{'}}

6 cos (0.2 t)

\int \frac{1}{x ( ln ( x ^{8} ) ) ^{9}} d x

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