What is the integral of (x^2+6x-8)/(x^3-4x) ?

The integral of (x^2+6x-8)/(x^3-4x) is 2ln|x|-2ln|x+2|+ln|x-2|+C

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integral of (x^2+6x-8)/(x^3-4x)

\int \frac{x ^{2} + 6 x - 8}{x ^{3} - 4 x} d x

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

Solution steps

\int \frac{x ^{2} + 6 x - 8}{x ^{3} - 4 x} d x

Take the partial fraction of

\frac{x ^{2} + 6 x - 8}{x ^{3} - 4 x} : \frac{2}{x} - \frac{2}{x + 2} + \frac{1}{x - 2}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (ln (3 x + y))

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

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

x \to 0 lim (\frac{4}{x} - \frac{4}{e ^{x} - 1})

\frac{\partial}{\partial x} ((6 x + 3 y) ey)

What is the integral of (x^2+6x-8)/(x^3-4x) ?
The integral of (x^2+6x-8)/(x^3-4x) is 2ln|x|-2ln|x+2|+ln|x-2|+C