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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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