What is the integral of (2x^3-4x^2-15x+5)/(x^2-2x-8) ?

The integral of (2x^3-4x^2-15x+5)/(x^2-2x-8) is x^2-1/2 ln|x+2|+3/2 ln|x-4|+C

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integral of (2x^3-4x^2-15x+5)/(x^2-2x-8)

\int \frac{2 x ^{3} - 4 x ^{2} - 15 x + 5}{x ^{2} - 2 x - 8} d x

x^{2} - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{3}{2} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{2 x ^{3} - 4 x ^{2} - 15 x + 5}{x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{2 x ^{3} - 4 x ^{2} - 15 x + 5}{x ^{2} - 2 x - 8} : 2 x - \frac{1}{2 ( x + 2 )} + \frac{3}{2 ( x - 4 )}

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

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

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

\int \frac{3}{2 ( x - 4 )} d x = \frac{3}{2} ln ∣ x - 4 ∣

= x^{2} - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{3}{2} ln ∣ x - 4 ∣

Add a constant to the solution

= x^{2} - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{3}{2} ln ∣ x - 4 ∣ + C

\frac{\partial}{\partial z} (x yz)

\frac{d}{d x} (cos (\frac{π}{4}))

\frac{d}{d x} (lo g_{10} (arcsin (cos (x))))

\int \frac{sin ( 2 x )}{15 + cos ^{2} ( x )} d x

\int \frac{10}{x ^{2} - 1} d x

What is the integral of (2x^3-4x^2-15x+5)/(x^2-2x-8) ?
The integral of (2x^3-4x^2-15x+5)/(x^2-2x-8) is x^2-1/2 ln|x+2|+3/2 ln|x-4|+C