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

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

Step 1 · Complete the square

$$ x^2+x+1 = \left(x+\frac12\right)^2 + \frac34. $$

Step 2 · Substitute

Let $u=x+\tfrac12$, $du=dx$:

$$ I = \int \frac{du}{u^2+\tfrac34} = \int \frac{du}{u^2+\left(\tfrac{\sqrt3}{2}\right)^2}. $$

Step 3 · Apply the standard arctangent formula

With $a=\dfrac{\sqrt3}{2}$:

$$ I = \frac{1}{a}\arctan\frac{u}{a} + C = \frac{2}{\sqrt3}\arctan\left(\frac{2u}{\sqrt3}\right) + C. $$

Step 4 · Substitute back

With $u=x+\tfrac12$, so $2u=2x+1$:

$$ I = \frac{2}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right) + C = \frac{2\sqrt3}{3}\arctan\left(\frac{2x+1}{\sqrt3}\right) + C. $$

Summary Table

Additional Notes

• Correction: the source rewrote $u^2+\tfrac34$ as $\tfrac34\!\left(\tfrac43u^2+1\right)$ and then multiplied the extracted factor $\tfrac43$ together with the standard formula's own $\tfrac{2}{\sqrt3}$ coefficient, effectively double-counting and producing $\tfrac{8}{3\sqrt3}$ instead of the correct $\tfrac{2}{\sqrt3}$. Applying $\int du/(u^2+a^2)=(1/a)\arctan(u/a)$ directly with $a=\sqrt3/2$ avoids this and gives the coefficient $2/\sqrt3=2\sqrt3/3$.
• The integrand is defined for all real $x$, since $x^2+x+1>0$ everywhere (its discriminant $1-4=-3$ is negative).
• The result can be checked by differentiating $\tfrac{2}{\sqrt3}\arctan\!\left(\tfrac{2x+1}{\sqrt3}\right)$: this gives $\tfrac{2}{\sqrt3}\cdot\tfrac{2/\sqrt3}{1+(2x+1)^2/3}=\tfrac{4/3}{[3+(2x+1)^2]/3}=\tfrac{4}{3+(2x+1)^2}=\tfrac{4}{4x^2+4x+4}=\tfrac{1}{x^2+x+1}$, matching the integrand.