1. **State the problem:** We want to evaluate the integral $$\int \frac{1}{x^4 + x^2 + 1} \, dx.$$\n\n2. **Rewrite the integrand:** Notice that the denominator can be seen as a quadratic in terms of $x^2$: $$x^4 + x^2 + 1 = (x^2)^2 + x^2 + 1.$$\n\n3. **Substitution:** Let $$u = x^2.$$ Then $$du = 2x \, dx,$$ but since there is no $x$ in the numerator, substitution directly with $u$ is not straightforward. Instead, we try to factor the denominator.\n\n4. **Factor the denominator:** We look for factors of the form $$(x^2 + ax + 1)(x^2 + bx + 1) = x^4 + (a+b)x^3 + (ab + 2)x^2 + (a+b)x + 1.$$\nMatching coefficients with $x^4 + x^2 + 1$, we get:\n- Coefficient of $x^3$: $a + b = 0$\n- Coefficient of $x^2$: $ab + 2 = 1$ so $ab = -1$\n- Coefficient of $x$: $a + b = 0$ (same as above)\n\nFrom $a + b = 0$, we have $b = -a$. Substitute into $ab = -1$:\n$$a(-a) = -1 \implies -a^2 = -1 \implies a^2 = 1 \implies a = \pm 1.$$\n\n5. **Choose $a = 1$, $b = -1$:** Then denominator factors as $$ (x^2 + x + 1)(x^2 - x + 1).$$\n\n6. **Partial fraction decomposition:** Write\n$$\frac{1}{(x^2 + x + 1)(x^2 - x + 1)} = \frac{Ax + B}{x^2 + x + 1} + \frac{Cx + D}{x^2 - x + 1}.$$\nMultiply both sides by the denominator:\n$$1 = (Ax + B)(x^2 - x + 1) + (Cx + D)(x^2 + x + 1).$$\n\n7. **Expand and collect terms:**\n$$(Ax + B)(x^2 - x + 1) = A x^3 - A x^2 + A x + B x^2 - B x + B,$$\n$$(Cx + D)(x^2 + x + 1) = C x^3 + C x^2 + C x + D x^2 + D x + D.$$\nSum:\n$$1 = (A + C) x^3 + (-A + B + C + D) x^2 + (A - B + C + D) x + (B + D).$$\n\n8. **Equate coefficients to zero except constant term:**\n- Coefficient of $x^3$: $A + C = 0$\n- Coefficient of $x^2$: $-A + B + C + D = 0$\n- Coefficient of $x$: $A - B + C + D = 0$\n- Constant term: $B + D = 1$\n\n9. **Solve the system:** From $A + C = 0$, $C = -A$.\nSubstitute into the other equations:\n- $-A + B - A + D = 0 \implies B + D - 2A = 0$\n- $A - B - A + D = 0 \implies -B + D = 0 \implies D = B$\n- $B + D = 1 \implies B + B = 1 \implies 2B = 1 \implies B = \frac{1}{2}$\nThen $D = \frac{1}{2}$.\nSubstitute back into $B + D - 2A = 0$:\n$$\frac{1}{2} + \frac{1}{2} - 2A = 0 \implies 1 - 2A = 0 \implies A = \frac{1}{2}.$$\nThen $C = -A = -\frac{1}{2}$.\n\n10. **Rewrite integral:**\n$$\int \frac{1}{x^4 + x^2 + 1} \, dx = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2 + x + 1} \, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2 - x + 1} \, dx.$$\n\n11. **Integrate each term separately:**\nFor the first integral, complete the square in denominator:\n$$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.$$\nLet $u = x + \frac{1}{2}$, then $du = dx$.\nRewrite numerator:\n$$\frac{1}{2}x + \frac{1}{2} = \frac{1}{2}(x + 1) = \frac{1}{2}(u + \frac{1}{2} + 1) = \frac{1}{2}u + \frac{3}{4}.$$\n\n12. **Split the first integral:**\n$$\int \frac{\frac{1}{2}u + \frac{3}{4}}{u^2 + \frac{3}{4}} \, du = \frac{1}{2} \int \frac{u}{u^2 + \frac{3}{4}} \, du + \frac{3}{4} \int \frac{1}{u^2 + \frac{3}{4}} \, du.$$\n\n13. **Integrate each part:**\n- For $$\int \frac{u}{u^2 + a^2} du = \frac{1}{2} \ln(u^2 + a^2) + C,$$ here $a^2 = \frac{3}{4}$.\n- For $$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan \frac{u}{a} + C.$$\n\n14. **Calculate first integral:**\n$$\frac{1}{2} \times \frac{1}{2} \ln \left(u^2 + \frac{3}{4}\right) + \frac{3}{4} \times \frac{2}{\sqrt{3}} \arctan \left( \frac{2u}{\sqrt{3}} \right) = \frac{1}{4} \ln \left(u^2 + \frac{3}{4}\right) + \frac{3}{2\sqrt{3}} \arctan \left( \frac{2u}{\sqrt{3}} \right).$$\n\n15. **Similarly for the second integral:**\nComplete the square:\n$$x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}.$$\nLet $v = x - \frac{1}{2}$, then numerator:\n$$-\frac{1}{2}x + \frac{1}{2} = -\frac{1}{2}(x - 1) = -\frac{1}{2}(v + \frac{1}{2} - 1) = -\frac{1}{2}v + \frac{1}{4}.$$\n\n16. **Split the second integral:**\n$$\int \frac{-\frac{1}{2}v + \frac{1}{4}}{v^2 + \frac{3}{4}} \, dv = -\frac{1}{2} \int \frac{v}{v^2 + \frac{3}{4}} \, dv + \frac{1}{4} \int \frac{1}{v^2 + \frac{3}{4}} \, dv.$$\n\n17. **Integrate each part:**\n- $$-\frac{1}{2} \times \frac{1}{2} \ln \left(v^2 + \frac{3}{4}\right) + \frac{1}{4} \times \frac{2}{\sqrt{3}} \arctan \left( \frac{2v}{\sqrt{3}} \right) = -\frac{1}{4} \ln \left(v^2 + \frac{3}{4}\right) + \frac{1}{2\sqrt{3}} \arctan \left( \frac{2v}{\sqrt{3}} \right).$$\n\n18. **Combine results and substitute back:**\n$$\int \frac{1}{x^4 + x^2 + 1} \, dx = \frac{1}{4} \ln \left(\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right) + \frac{3}{2\sqrt{3}} \arctan \left( \frac{2x + 1}{\sqrt{3}} \right) - \frac{1}{4} \ln \left(\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}\right) + \frac{1}{2\sqrt{3}} \arctan \left( \frac{2x - 1}{\sqrt{3}} \right) + C.$$\n\nThis is the final answer.