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

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

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

\int \frac{2 x ^{3} - 4 x - 8}{( x ^{2} - x ) ( x ^{2} + 4 )} d x

2 ln ∣ x ∣ - 2 ln ∣ x - 1 ∣ + ln x^{2} + 4 + 2 arctan (\frac{x}{2}) + C

Solution steps

\int \frac{2 x ^{3} - 4 x - 8}{( x ^{2} - x ) ( x ^{2} + 4 )} d x

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

\int \frac{2 x + 4}{x ^{2} + 4} d x = ln x^{2} + 4 + 2 arctan (\frac{x}{2})

= 2 ln ∣ x ∣ - 2 ln ∣ x - 1 ∣ + ln x^{2} + 4 + 2 arctan (\frac{x}{2})

Add a constant to the solution

= 2 ln ∣ x ∣ - 2 ln ∣ x - 1 ∣ + ln x^{2} + 4 + 2 arctan (\frac{x}{2}) + C

0.0588 (80)^{1.125}

\frac{\partial}{\partial x} (\frac{2 x}{2 + y})

x y^{^{'}} - 2 y = x^{2} - x - 1

\frac{s}{( s ^{3} + 1 )}

\int_{a}^{b} \frac{1}{x} d x

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