What is the integral of (x+2)/(x^2-3x-4) ?

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

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integral of (x+2)/(x^2-3x-4)

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

- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{6}{5} ln ∣ x - 4 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{5 ( x + 1 )} + \frac{6}{5 ( 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}{5 ( x + 1 )} d x + \int \frac{6}{5 ( x - 4 )} d x

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

\int \frac{6}{5 ( x - 4 )} d x = \frac{6}{5} ln ∣ x - 4 ∣

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{6}{5} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{6}{5} ln ∣ x - 4 ∣ + C

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