Evaluate the integral: I = integral dx/((x²+x+1)
Evaluate the integral: $$ I = \int \frac{dx}{\left(x^2+x+1\right)\sqrt{x^2+x+1}} = \int\frac{dx}{\left(x^2+x+1\right)^{3/2}}. $$
Explanation of Method
Completing the square and substituting $u=\tfrac{\sqrt3}2\tan\theta$ collapses $\left(u^2+\tfrac34\right)^{3/2}$ against the $\sec^2\theta\,d\theta$ from $du$, reducing the integral to $\int\cos\theta\,d\theta$.
- Complete the square and substitute $u=x+\tfrac12$.
- Apply the trig substitution $u=\tfrac{\sqrt3}2\tan\theta$.
- Integrate $\cos\theta$.
- Convert back to $x$.
Step 1 · Complete the square
$x^2+x+1=\left(x+\tfrac12\right)^2+\tfrac34$. Let $u=x+\tfrac12$:
$$ I = \int\frac{du}{\left(u^2+3/4\right)^{3/2}}. $$
Step 2 · Trig substitution
Let $u=\dfrac{\sqrt3}2\tan\theta$, so $du=\dfrac{\sqrt3}2\sec^2\theta\,d\theta$ and $u^2+\tfrac34=\tfrac34\sec^2\theta$:
$$ I = \int\frac{\frac{\sqrt3}2\sec^2\theta\,d\theta}{\left(\frac34\right)^{3/2}\sec^3\theta} = \frac{\sqrt3/2}{(3\sqrt3)/8}\int\cos\theta\,d\theta = \frac43\int\cos\theta\,d\theta. $$
Step 3 · Integrate
$$ I = \frac43\sin\theta + C. $$
Step 4 · Convert back to $x$
From $\tan\theta=2u/\sqrt3$, $\sin\theta=\dfrac{\tan\theta}{\sqrt{1+\tan^2\theta}}=\dfrac{2u}{\sqrt{3+4u^2}}$; and $3+4u^2=4\left(x^2+x+1\right)$, so $\sqrt{3+4u^2}=2\sqrt{x^2+x+1}$:
$$ I = \frac43\cdot\frac{2\left(x+\frac12\right)}{2\sqrt{x^2+x+1}} + C = \frac{2(2x+1)}{3\sqrt{x^2+x+1}} + C. $$
Summary Table
| Substitution | Effect |
|---|---|
| $u=x+1/2$ | $x^2+x+1\to u^2+3/4$ |
| $u=\tfrac{\sqrt3}2\tan\theta$ | reduces to $\tfrac43\int\cos\theta\,d\theta$ |
Additional Notes
- The domain is all real $x$, since $x^2+x+1>0$ everywhere.
- The result was verified by differentiation at $x=0$ and $x=1$: the derivative matches the integrand $1/(x^2+x+1)^{3/2}$ exactly at both points.
Result
The solution to the integral is: