Evaluate the integral: I = integral (x³)/(x²+x+1) dx

Evaluate the integral: $$ I = \int \frac{x^3}{x^2+x+1}\,dx. $$

Step 1 · Polynomial long division

$$ \frac{x^3}{x^2+x+1} = x-1+\frac{1}{x^2+x+1}. $$

Step 2 · Integrate the polynomial part

$$ \int (x-1)\,dx = \frac{x^2}{2} - x. $$

Step 3 · Integrate the remainder

Completing the square, $x^2+x+1=(x+\tfrac12)^2+\tfrac34$, and using the standard arctangent formula:

$$ \int \frac{dx}{x^2+x+1} = \frac{2}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right) + C. $$

Step 4 · Combine

$$ I = \frac{x^2}{2} - x + \frac{2}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right) + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $x^2+x+1>0$ everywhere.
• The arctangent coefficient here is applied directly from the standard formula $\int du/(u^2+a^2)=(1/a)\arctan(u/a)$ with $a=\sqrt3/2$, giving $2/\sqrt3$ — this is the same result independently verified (after correcting an unrelated intermediate slip in the original derivation) in Problem 189.
• The result can be checked by differentiating $\tfrac{x^2}2-x+\tfrac2{\sqrt3}\arctan\!\left(\tfrac{2x+1}{\sqrt3}\right)$: this gives $x-1+\tfrac{1}{x^2+x+1}=\tfrac{(x-1)(x^2+x+1)+1}{x^2+x+1}=\tfrac{x^3-1+1}{x^2+x+1}=\tfrac{x^3}{x^2+x+1}$, matching the integrand.