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

The integral of x/(x^3+1) is -1/3 ln|x+1|+1/6 (ln|4x^2-4x+4|+2sqrt(3)arctan((2x-1)/(sqrt(3))))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(x^3+1)

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

- \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3})) + C

Solution steps

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

Take the partial fraction of

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

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

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

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

\int \frac{x + 1}{3 ( x ^{2} - x + 1 )} d x = \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3}))

= - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3}))

Add a constant to the solution

= - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (ln 4 x^{2} - 4 x + 4 + 23 arctan (\frac{2 x - 1}{3})) + C

\int sin (x) + sin (x) cos (x) d x

\int \frac{4 x}{x ^{3} + 1} d x

\int (4 x^{2} + 1)^{3} \cdot 8 x d x

x (1 - x^{2})

x \to 2 lim (x x^{2} - 4)

What is the integral of x/(x^3+1) ?
The integral of x/(x^3+1) is -1/3 ln|x+1|+1/6 (ln|4x^2-4x+4|+2sqrt(3)arctan((2x-1)/(sqrt(3))))+C