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

The integral of 1/(x^2-x-6) is -1/5 ln|x+2|+1/5 ln|x-3|+C

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

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

- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{5 ( x + 2 )} + \frac{1}{5 ( x - 3 )} 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}{5 ( x + 2 )} d x + \int \frac{1}{5 ( x - 3 )} d x

\int \frac{1}{5 ( x + 2 )} d x = \frac{1}{5} ln ∣ x + 2 ∣

\int \frac{1}{5 ( x - 3 )} d x = \frac{1}{5} ln ∣ x - 3 ∣

= - \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ + C

\frac{\partial}{\partial x} (- \frac{y}{x})

f (x) = \frac{( x )}{x - 4}

\frac{\partial}{\partial x} (sin (x^{2} + 5))

(8 - r^{2}) d r = r^{3} sin (θ) d θ

\int_{0}^{1} \frac{x - 6}{x ^{2} - 6 x + 8} d x

What is the integral of 1/(x^2-x-6) ?
The integral of 1/(x^2-x-6) is -1/5 ln|x+2|+1/5 ln|x-3|+C