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

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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

= 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

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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