Evaluate the integral: I = integral dx/(x²sqrt(x²-x+1))

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

Step 1 · Substitute

Let $u=1/x$, so $x=1/u$, $dx=-du/u^2$, and (for $x>0$) $\sqrt{x^2-x+1}=\sqrt{1-u+u^2}/u$:

$$ I = -\int\frac{u\,du}{\sqrt{1-u+u^2}}. $$

Step 2 · Complete the square and split

Writing $u=(v+\tfrac12)$ with $v=u-\tfrac12$, so $1-u+u^2=v^2+\tfrac34$:

$$ -\int\frac{v+\tfrac12}{\sqrt{v^2+3/4}}\,dv = -\int\frac{v\,dv}{\sqrt{v^2+3/4}} - \frac12\int\frac{dv}{\sqrt{v^2+3/4}}. $$

Step 3 · Integrate

$$ = -\sqrt{v^2+\tfrac34} - \frac12\ln\left|v+\sqrt{v^2+\tfrac34}\right| + C = -\sqrt{u^2-u+1} - \frac12\ln\left|u-\frac12+\sqrt{u^2-u+1}\right| + C. $$

Step 4 · Back-substitute

With $u=1/x$ (for $x>0$), $\sqrt{u^2-u+1}=\sqrt{x^2-x+1}/x$ and $u-\tfrac12=(2-x)/(2x)$:

$$ I = -\frac{\sqrt{x^2-x+1}}x - \frac12\ln\left|\frac{2-x+2\sqrt{x^2-x+1}}{2x}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$.
• The result was verified numerically at $x=1$ and $x=2$: the derivative evaluates to $1$ and $\approx0.1443$ respectively, matching the integrand values $1/[1\cdot1]=1$ and $1/[4\sqrt3]\approx0.1443$ exactly.