Evaluate the integral: I = integral (x+1)/(x²+x+1) dx
Evaluate the integral: $$ I = \int \frac{x+1}{x^2+x+1}\,dx. $$
Explanation of Method
Writing the numerator as half the derivative of the denominator plus a constant splits the integral into a logarithm and a standard arctangent integral.
- Rewrite $x+1=\tfrac12(2x+1)+\tfrac12$.
- Split into a derivative-matching piece and a constant piece.
- Integrate each piece.
Step 1 · Rewrite the numerator
Since $\dfrac{d}{dx}(x^2+x+1)=2x+1$, write:
$$ x+1 = \frac12(2x+1) + \frac12. $$
Step 2 · Split the integral
$$ I = \frac12\int \frac{2x+1}{x^2+x+1}\,dx + \frac12\int \frac{dx}{x^2+x+1}. $$
Step 3 · Integrate each piece
The first integral is a direct log-derivative match:
$$ \frac12\int \frac{2x+1}{x^2+x+1}\,dx = \frac12\ln\left(x^2+x+1\right). $$
The second integral is the standard arctangent result (Problem 189):
$$ \int \frac{dx}{x^2+x+1} = \frac{2}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right), \qquad\text{so}\qquad \frac12\int \frac{dx}{x^2+x+1} = \frac{1}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right). $$
Step 4 · Combine
$$ I = \frac12\ln\left(x^2+x+1\right) + \frac{1}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right) + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $(2x+1)/(x^2+x+1)$ | $\ln(x^2+x+1)$ |
| $1/(x^2+x+1)$ | $\tfrac{2}{\sqrt3}\arctan\!\left(\tfrac{2x+1}{\sqrt3}\right)$ |
Additional Notes
- The integrand is defined for all real $x$, since $x^2+x+1>0$ everywhere.
- Since $x^2+x+1>0$ always, the absolute value bars around the log argument can be dropped.
- The result can be checked by differentiating $\tfrac12\ln(x^2+x+1)+\tfrac1{\sqrt3}\arctan\!\left(\tfrac{2x+1}{\sqrt3}\right)$: this gives $\tfrac{2x+1}{2(x^2+x+1)}+\tfrac{1}{2(x^2+x+1)}=\tfrac{2x+2}{2(x^2+x+1)}=\tfrac{x+1}{x^2+x+1}$, matching the integrand.
Result
The solution to the integral is: