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

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

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

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

ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

= ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

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