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

The integral of (4x-1)/(x^2-x-2) is 5/3 ln|x+1|+7/3 ln|x-2|+C

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

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

\frac{5}{3} ln ∣ x + 1 ∣ + \frac{7}{3} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{4 x - 1}{x ^{2} - x - 2} : \frac{5}{3 ( x + 1 )} + \frac{7}{3 ( x - 2 )}

= \int \frac{5}{3 ( x + 1 )} + \frac{7}{3 ( x - 2 )} 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{5}{3 ( x + 1 )} d x + \int \frac{7}{3 ( x - 2 )} d x

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

\int \frac{7}{3 ( x - 2 )} d x = \frac{7}{3} ln ∣ x - 2 ∣

= \frac{5}{3} ln ∣ x + 1 ∣ + \frac{7}{3} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{5}{3} ln ∣ x + 1 ∣ + \frac{7}{3} ln ∣ x - 2 ∣ + C

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What is the integral of (4x-1)/(x^2-x-2) ?
The integral of (4x-1)/(x^2-x-2) is 5/3 ln|x+1|+7/3 ln|x-2|+C