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

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

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integral of x/(x^3-x)

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

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

Solution steps

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

Simplify

\frac{x}{x ^{3} - x} : \frac{1}{x ^{2} - 1}

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

Factor

x^{2} - 1 : - (- x^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{- 2 x ^{3}}{x ^{2} + 9} d x

\int \frac{sec ( x )}{cot ( x )} d x

\int (1 + x^{2}) cos (x) d x

\int \frac{1}{x ^{2} - c ^{2}} d x

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