What is the integral of (x+1)/(x^2-2x-8) ?

The integral of (x+1)/(x^2-2x-8) is 1/6 ln|x+2|+5/6 ln|x-4|+C

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integral of (x+1)/(x^2-2x-8)

\int \frac{x + 1}{x ^{2} - 2 x - 8} d x

\frac{1}{6} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x + 1}{x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{x + 1}{x ^{2} - 2 x - 8} : \frac{1}{6 ( x + 2 )} + \frac{5}{6 ( x - 4 )}

= \int \frac{1}{6 ( x + 2 )} + \frac{5}{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{1}{6 ( x + 2 )} d x + \int \frac{5}{6 ( x - 4 )} d x

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

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

= \frac{1}{6} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x + 2 ∣ + \frac{5}{6} ln ∣ x - 4 ∣ + C

\int \frac{4 x + 6}{( x ^{2} + 3 x + 7 ) ^{4}} d x

\int x^{2} (x - 1)^{4} d x

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

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

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

What is the integral of (x+1)/(x^2-2x-8) ?
The integral of (x+1)/(x^2-2x-8) is 1/6 ln|x+2|+5/6 ln|x-4|+C