What is the integral of 1/(x^2-4x-12) ?

The integral of 1/(x^2-4x-12) is -1/8 ln|x+2|+1/8 ln|x-6|+C

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integral of 1/(x^2-4x-12)

\int \frac{1}{x ^{2} - 4 x - 12} d x

- \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 4 x - 12} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 12} : - \frac{1}{8 ( x + 2 )} + \frac{1}{8 ( x - 6 )}

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

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

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

= - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 6 ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 6 ∣ + C

\frac{20}{s ( s ^{2} + 4 )}

\int \frac{1}{1 + cos ( 3 x )} d x

\frac{\partial}{\partial x} (22 x y e^{- x^{2} - y^{2}})

- 0.1 x^{2} + 1.2 x + 98.4

2 x y d x + (x^{2} + 2 y) d y = 0

What is the integral of 1/(x^2-4x-12) ?
The integral of 1/(x^2-4x-12) is -1/8 ln|x+2|+1/8 ln|x-6|+C