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

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

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

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

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

Solution steps

\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 )}

= \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

\int \frac{( 2 x + 1 )}{x ^{2} - 2 x + 5} d x

x \to \frac{π}{4} lim (tan^{2} (x))

\int 9 tan (x) d x

x \to - 1 lim (\frac{x + 2}{x + 1})

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