91Ó°ÊÓ

Let \(t = x^2\). Then, the differential of \(t\) is \(\mathrm{dt} = 2x\,\mathrm{dx}\). Now, we substitute \(t\) into the integral and simplify: $$ \int \frac{\mathrm{dx}}{x^4 + x^2 + 1} = \int \frac{2x\,\mathrm{dx}}{x^4 + x^2 + 1} = \int \frac{\mathrm{dt}}{t^2 + t + 1} $$

To decompose the fraction expression into simpler terms, first find the roots of the denominator \(t^2 + t + 1\). We can use the quadratic formula to find \(t = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}\). These roots are complex numbers: \(t_1 = \frac{-1 + \sqrt{-3}}{2}\) and \(t_2 = \frac{-1 - \sqrt{-3}}{2}\). Let \(A\) and \(B\) be coefficients for \(t_1\) and \(t_2\) respectively, and express the fraction as a sum of partial fractions: $$ \frac{1}{t^2 + t + 1} = \frac{A}{t-t_1} + \frac{B}{t-t_2} $$ Now, clear the denominator by multiplying both sides by \((t-t_1)(t-t_2)\): $$ 1 = A(t-t_2) + B(t-t_1) $$ Now we'll solve for \(A\) and \(B\). To find \(A\), let \(t = t_1\): \(1 = A(t_1 - t_2)\), so \(A = \frac{1}{t_1 - t_2}\). Similarly, to find \(B\), let \(t = t_2\): \(1 = B(t_2 - t_1)\), so \(B = \frac{1}{t_2 - t_1}\).

Plug the values of \(A\) and \(B\) back into the integral: $$ \int \frac{\mathrm{dt}}{t^2 + t + 1} = \int \left(\frac{A}{t-t_1} + \frac{B}{t-t_2}\right)\mathrm{dt} = \int \left(\frac{\frac{1}{t_1 - t_2}}{t-t_1} + \frac{\frac{1}{t_2 - t_1}}{t-t_2}\right)\mathrm{dt} $$ Now, proceed with the integration: $$ \int \frac{\mathrm{dt}}{t^2 + t + 1} = \frac{1}{t_1 - t_2}\log|t-t_1| - \frac{1}{t_2 - t_1}\log|t-t_2| + C $$ Finally, substitute back \(x^2\) for \(t\) and simplify: $$ \int \frac{\mathrm{dx}}{x^4 + x^2 + 1} = \frac{1}{t_1 - t_2}\log|x^2-t_1| - \frac{1}{t_2 - t_1}\log|x^2-t_2| + C $$