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

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

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

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

\frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{5 ( x + 2 )} + \frac{4}{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{4}{5 ( x - 3 )} d x

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

\int \frac{4}{5 ( x - 3 )} d x = \frac{4}{5} ln ∣ x - 3 ∣

= \frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣ + C

\frac{d}{d x} ((e^{x} + 1)^{7})

\int \frac{x - 3}{- x ^{2} + 4 x - 5} d x

\frac{\partial}{\partial y} (y e^{x y} - \frac{1}{y})

\frac{d}{d x} (4 x^{3} - 66 x^{2} + 216 x)

\frac{d}{d x} (cos^{2} (x^{3}) - ln (x^{2} - 1))

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