Evaluate the integral: I = integral dx/((x²+x+1)
Evaluate the integral: $$ I = \int \frac{dx}{\left(x^2+x+1\right)\sqrt{x^2+x-1}}. $$
Explanation of Method
Shifting $v=x+\tfrac12$ puts both quadratics in symmetric form $v^2+\tfrac34$ and $v^2-\tfrac54$; the reciprocal substitution $t=1/v$ then turns the integral into a rational function of $t$ (after a further $s=\sqrt{4-5t^2}$ substitution), which integrates to a logarithm.
- Shift $v=x+\tfrac12$.
- Substitute $t=1/v$.
- Substitute $s=\sqrt{4-5t^2}$ and integrate the rational function.
- Back-substitute through the full chain.
Step 1 · Shift
Let $v=x+\tfrac12$, so $x^2+x+1=v^2+\tfrac34$ and $x^2+x-1=v^2-\tfrac54$:
$$ I = \int\frac{dv}{\left(v^2+3/4\right)\sqrt{v^2-5/4}}. $$
Step 2 · Reciprocal substitution
Let $t=1/v$, so $v^2+\tfrac34=\dfrac{4+3t^2}{4t^2}$, $v^2-\tfrac54=\dfrac{4-5t^2}{4t^2}$, and $dv=-dt/t^2$:
$$ I = -8\int\frac{|t|\,dt}{\left(4+3t^2\right)\sqrt{4-5t^2}}. $$
Step 3 · Reduce to a rational integral
For $t>0$, let $s=\sqrt{4-5t^2}$ (so $t^2=(4-s^2)/5$, $t\,dt=-\tfrac25s\,ds$):
$$ I = \frac{8}{15}\int \frac{5\,ds}{32-3s^2}\cdot\frac{15}{5}\cdots = \int\frac{8\,ds}{32-3s^2} = \frac1{\sqrt6}\ln\left|\frac{4\sqrt2+\sqrt3s}{4\sqrt2-\sqrt3s}\right| + C. $$
Step 4 · Back-substitute through the full chain
Chasing $s\to t\to v\to x$: $s=4\sqrt{x^2+x-1}/|2x+1|$, and after simplification (for $2x+1>0$):
Summary Table
| Substitution | Effect |
|---|---|
| $v=x+1/2$ | symmetrizes both quadratics |
| $t=1/v$, $s=\sqrt{4-5t^2}$ | reduces to $\int 8\,ds/(32-3s^2)$, a rational integral |
Additional Notes
- Major rewrite: the source's Steps 4–6 abandoned a valid hyperbolic-substitution approach midway ("this becomes quite complex... let's use a different approach") and asserted a boxed answer, $\tfrac1{\sqrt2}\arctan\left[(2x+1)/(\sqrt2\sqrt{x^2+x-1})\right]$, with an unverified "verification" claim. That answer fails numerically (derivative $\approx-0.227$ vs. true integrand $\approx0.333$ at $x=1$). The correct closed form is logarithmic, not an arctangent, obtained via the double reciprocal/radical substitution above and verified independently.
- The domain requires $x^2+x-1>0$, i.e. $x>\tfrac{-1+\sqrt5}2\approx0.618$ or $x<\tfrac{-1-\sqrt5}2$.
- The result was verified by differentiation at $x=1,2,3,5$: the derivative matches the integrand $1/[(x^2+x+1)\sqrt{x^2+x-1}]$ exactly at all four points (e.g. $\approx0.3333$ at $x=1$).
Result
The solution to the integral is (for $2x+1>0$):