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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

integral

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

- 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

- (\frac{x}{( 1 - x )})

2^{10 x - 1}

\frac{x ^{- 2} ( e ^{x} ) ^{3}}{3}

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

4 x^{3} + 3 x^{2} + 2

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