Evaluate the integral: I = integral dx/((x²+x+1)³)
Evaluate the integral: $$ I = \int \frac{dx}{\left(x^2+x+1\right)^3}. $$
Explanation of Method
Completing the square reduces the denominator to a shifted, scaled version of $(v^2+1)^3$, allowing the standard reduction formula for $\int dv/(v^2+1)^n$ to be applied twice.
- Complete the square in the denominator.
- Substitute to normalize to $(v^2+1)^3$.
- Apply the reduction formula twice.
- Back-substitute and simplify.
Step 1 · Complete the square and normalize
$$ x^2+x+1=\left(x+\frac12\right)^2+\frac34. $$
Let $v=\dfrac{2x+1}{\sqrt3}$, so that $x^2+x+1=\dfrac34(v^2+1)$ and $dx=\dfrac{\sqrt3}{2}dv$:
$$ I = \int\frac{dx}{\left(\tfrac34\right)^3(v^2+1)^3} = \frac{64}{27}\cdot\frac{\sqrt3}2\int\frac{dv}{(v^2+1)^3} = \frac{32\sqrt3}{27}\int\frac{dv}{(v^2+1)^3}. $$
Step 2 · Apply the reduction formula twice
Using $\displaystyle\int\frac{dv}{(v^2+1)^n}=\frac{v}{2(n-1)(v^2+1)^{n-1}}+\frac{2n-3}{2(n-1)}\int\frac{dv}{(v^2+1)^{n-1}}$:
$$ \int\frac{dv}{(v^2+1)^2} = \frac{v}{2(v^2+1)}+\frac12\arctan v, $$
$$ \int\frac{dv}{(v^2+1)^3} = \frac{v}{4(v^2+1)^2} + \frac34\int\frac{dv}{(v^2+1)^2} = \frac{v}{4(v^2+1)^2} + \frac{3v}{8(v^2+1)} + \frac38\arctan v. $$
Step 3 · Back-substitute
Since $v=\dfrac{2x+1}{\sqrt3}$ and $v^2+1=\dfrac{4(x^2+x+1)}3$:
$$ \frac{32\sqrt3}{27}\cdot\frac{v}{4(v^2+1)^2} = \frac{2x+1}{6(x^2+x+1)^2}, \qquad \frac{32\sqrt3}{27}\cdot\frac{3v}{8(v^2+1)} = \frac{2x+1}{3(x^2+x+1)}, $$
$$ \frac{32\sqrt3}{27}\cdot\frac38\arctan v = \frac{4\sqrt3}9\arctan\left(\frac{2x+1}{\sqrt3}\right). $$
Summary Table
| Piece | Contribution to $I$ |
|---|---|
| $v/[4(v^2+1)^2]$ term | $(2x+1)/[6(x^2+x+1)^2]$ |
| $3v/[8(v^2+1)]$ term | $(2x+1)/[3(x^2+x+1)]$ |
| $\tfrac38\arctan v$ term | $(4\sqrt3/9)\arctan\left(\tfrac{2x+1}{\sqrt3}\right)$ |
Additional Notes
- The domain is all real $x$, since $x^2+x+1>0$ everywhere.
- The result was verified by direct differentiation: combining the derivatives of all three terms over a common denominator collapses to exactly $1/(x^2+x+1)^3$, matching the integrand.
Result
The solution to the integral is: