What is the integral of (x^3-11x-26)/(x^2-2x-8) ?

The integral of (x^3-11x-26)/(x^2-2x-8) is (x^2)/2+2x+2ln|x+2|-ln|x-4|+C

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integral of (x^3-11x-26)/(x^2-2x-8)

\int \frac{x ^{3} - 11 x - 26}{x ^{2} - 2 x - 8} d x

\frac{x ^{2}}{2} + 2 x + 2 ln ∣ x + 2 ∣ - ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x ^{3} - 11 x - 26}{x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{x ^{3} - 11 x - 26}{x ^{2} - 2 x - 8} : x + 2 + \frac{2}{x + 2} - \frac{1}{x - 4}

= \int x + 2 + \frac{2}{x + 2} - \frac{1}{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 x d x + \int 2 d x + \int \frac{2}{x + 2} d x - \int \frac{1}{x - 4} d x

\int x d x = \frac{x ^{2}}{2}

\int 2 d x = 2 x

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

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

= \frac{x ^{2}}{2} + 2 x + 2 ln ∣ x + 2 ∣ - ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + 2 ln ∣ x + 2 ∣ - ln ∣ x - 4 ∣ + C

\int \frac{1}{sin ^{4} ( 3 x ) cos ^{2} ( 3 x )} d x

\frac{x ^{2}}{x ^{2} - 1}

x \to 0 lim (\frac{6}{x ^{2}})

cos (θ) - θ sin (θ)

\frac{d}{d x} (\frac{a x + b}{c x - a})

What is the integral of (x^3-11x-26)/(x^2-2x-8) ?
The integral of (x^3-11x-26)/(x^2-2x-8) is (x^2)/2+2x+2ln|x+2|-ln|x-4|+C