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

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

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

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

2 ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

= 2 ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= 2 ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial x} (7 sin (x) sin (y))

\frac{d}{d x} (\frac{3}{4})

\frac{d}{d x} (\frac{6}{5})

\frac{d}{d v} (\frac{v}{t})

\frac{d}{d t} (6 arctan (t))

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