What is the integral of (x^2+12x-1)/(x^3-x) ?

The integral of (x^2+12x-1)/(x^3-x) is ln|x|-6ln|x+1|+6ln|x-1|+C

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integral of (x^2+12x-1)/(x^3-x)

\int \frac{x ^{2} + 12 x - 1}{x ^{3} - x} d x

ln ∣ x ∣ - 6 ln ∣ x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{2} + 12 x - 1}{x ^{3} - x} d x

Take the partial fraction of

\frac{x ^{2} + 12 x - 1}{x ^{3} - x} : \frac{1}{x} - \frac{6}{x + 1} + \frac{6}{x - 1}

= \int \frac{1}{x} - \frac{6}{x + 1} + \frac{6}{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{6}{x + 1} d x + \int \frac{6}{x - 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

\int \frac{6}{x - 1} d x = 6 ln ∣ x - 1 ∣

= ln ∣ x ∣ - 6 ln ∣ x + 1 ∣ + 6 ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - 6 ln ∣ x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

\int \frac{- sin ( x ) tan ( x )}{1} d x

\frac{d}{d x} (\frac{14}{x})

\frac{6}{3 x + 1}, at x = - 1

\frac{\partial}{\partial x} (- x - 6)

\frac{\partial}{\partial x} (4 - x^{2} - y^{2})

What is the integral of (x^2+12x-1)/(x^3-x) ?
The integral of (x^2+12x-1)/(x^3-x) is ln|x|-6ln|x+1|+6ln|x-1|+C