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

The integral of x/((1+x^3)) 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/((1+x^3))

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

Take the partial fraction of

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

x \to 0 + lim (sin (x))

\int (a x^{3} - b x^{2} - c x + d) d x

\int_{- 2}^{4} ∣ x ∣ d x

\frac{d y}{d x} = x e^{y}

y^{^{''}} - 4 y^{^{'}} + \frac{25}{4} y = 10 cos (8 x)

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