What is the integral of (2x-7)/(x^2-2x-8) ?

The integral of (2x-7)/(x^2-2x-8) is 11/6 ln|x+2|+1/6 ln|x-4|+C

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integral of (2x-7)/(x^2-2x-8)

\int \frac{2 x - 7}{x ^{2} - 2 x - 8} d x

\frac{11}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{2 x - 7}{x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{2 x - 7}{x ^{2} - 2 x - 8} : \frac{11}{6 ( x + 2 )} + \frac{1}{6 ( x - 4 )}

= \int \frac{11}{6 ( x + 2 )} + \frac{1}{6 ( 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{11}{6 ( x + 2 )} d x + \int \frac{1}{6 ( x - 4 )} d x

\int \frac{11}{6 ( x + 2 )} d x = \frac{11}{6} ln ∣ x + 2 ∣

\int \frac{1}{6 ( x - 4 )} d x = \frac{1}{6} ln ∣ x - 4 ∣

= \frac{11}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{11}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣ + C

\int (1 - 3 x) 6 x^{5} d x

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

\int - 9 x^{4} e^{x^{5}} d x

\int \frac{1}{( 1 + sin ( x ) + cos ( x ) )} d x

\frac{1}{2 x}

What is the integral of (2x-7)/(x^2-2x-8) ?
The integral of (2x-7)/(x^2-2x-8) is 11/6 ln|x+2|+1/6 ln|x-4|+C