What is the integral of 3/(x^3-1) ?

The integral of 3/(x^3-1) is ln|x-1|-1/2 ln|4x^2+4x+4|-sqrt(3)arctan(1/(sqrt(3))(2x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 3/(x^3-1)

\int \frac{3}{x ^{3} - 1} d x

ln ∣ x - 1 ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - 3 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

\int \frac{3}{x ^{3} - 1} d x

Take the constant out:

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

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

Take the partial fraction of

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

= 3 \cdot \int \frac{1}{3 ( x - 1 )} + \frac{- x - 2}{3 ( x ^{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

= 3 (\int \frac{1}{3 ( x - 1 )} d x + \int \frac{- x - 2}{3 ( x ^{2} + x + 1 )} d x)

\int \frac{1}{3 ( x - 1 )} d x = \frac{1}{3} ln ∣ x - 1 ∣

\int \frac{- x - 2}{3 ( x ^{2} + x + 1 )} d x = \frac{1}{6 3} (- 3 ln 4 x^{2} + 4 x + 4 - 6 arctan (\frac{2 x + 1}{3}))

= 3 (\frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6 3} (- 3 ln 4 x^{2} + 4 x + 4 - 6 arctan (\frac{2 x + 1}{3})))

Simplify

3 (\frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6 3} (- 3 ln 4 x^{2} + 4 x + 4 - 6 arctan (\frac{2 x + 1}{3}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - 3 arctan (\frac{1}{3} (2 x + 1))

= ln ∣ x - 1 ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - 3 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{1}{2} ln 4 x^{2} + 4 x + 4 - 3 arctan (\frac{1}{3} (2 x + 1)) + C

What is the integral of 3/(x^3-1) ?
The integral of 3/(x^3-1) is ln|x-1|-1/2 ln|4x^2+4x+4|-sqrt(3)arctan(1/(sqrt(3))(2x+1))+C