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

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

Popular >

integral of (x+2)/(x^3-1)

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

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

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{x - 1} + \frac{- x - 1}{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

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

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

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

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

\int x sin (2 x + 3) d x

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

\int csc (θ) (csc (θ) + cot (θ)) d θ

\int \frac{8}{5 - 2 x} d x

\int \frac{5 x ^{3}}{x ^{2} + 1} d x

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