What is the integral of (x-4)/(x^2-x-6) ?

The integral of (x-4)/(x^2-x-6) is 6/5 ln|x+2|-1/5 ln|x-3|+C

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integral of (x-4)/(x^2-x-6)

\int \frac{x - 4}{x ^{2} - x - 6} d x

\frac{6}{5} ln ∣ x + 2 ∣ - \frac{1}{5} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x - 4}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{x - 4}{x ^{2} - x - 6} : \frac{6}{5 ( x + 2 )} - \frac{1}{5 ( x - 3 )}

= \int \frac{6}{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{6}{5 ( x + 2 )} d x - \int \frac{1}{5 ( x - 3 )} d x

\int \frac{6}{5 ( x + 2 )} d x = \frac{6}{5} ln ∣ x + 2 ∣

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

= \frac{6}{5} ln ∣ x + 2 ∣ - \frac{1}{5} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{6}{5} ln ∣ x + 2 ∣ - \frac{1}{5} ln ∣ x - 3 ∣ + C

2 x d y + y d x = 0

\int \frac{4}{x ^{2} + 2 x} d x

\frac{\partial}{\partial x} (x ln (z^{5} + x^{7}))

x y y^{^{'}} + 4 x^{2} + y^{2} = 0

y^{^{''}} + 6 y^{^{'}} + 10 = 0

What is the integral of (x-4)/(x^2-x-6) ?
The integral of (x-4)/(x^2-x-6) is 6/5 ln|x+2|-1/5 ln|x-3|+C