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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= - 2 \cdot \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

= - 2 (- \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 ∣

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

Simplify

- 2 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) : 2 ln ∣ x ∣ - ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\int_{1}^{b} \frac{1}{( x + 4 ) ^{2}} d x

\int \frac{1}{6} sin (6 t) d t

f (x) = 3 + x^{2}

y = x^{2} - x + 2, at x = 2

\frac{d}{d x} ((x + 3)^{2})

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