What is the integral of (x^2+8x-1)/(x^3-x) ?

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

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integral of (x^2+8x-1)/(x^3-x)

\int \frac{x ^{2} + 8 x - 1}{x ^{3} - x} d x

ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{2} + 8 x - 1}{x ^{3} - x} d x

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - 4 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ + C

x \to - 1 lim (2 x^{2} - \frac{2}{x - 1})

\frac{\partial}{\partial x} (x^{9} y^{4} - x^{6} + y^{8})

(- 5, 1), (- 4, - 7)

\frac{x}{e ^{x}}

n = 0 \sum \infty (- 1)^{n + 1}

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