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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply Integration By Parts

= x (x - 1) - \int x d x

\int x d x = \frac{x ^{2}}{2}

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

Add a constant to the solution

= x (x - 1) - \frac{x ^{2}}{2} + C

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