$\int \frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}$ is equal to

$\begin{array}{l}\int \frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^4+{\mathrm{x}}^2+1}\mathrm{dx}\\ \Rightarrow \mathrm{I}=\int \frac{1-\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+1}\mathrm{dx}=\int \frac{1-\frac{1}{{\mathrm{x}}^2}}{{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2-1}\mathrm{dx}\\ \text{ Put }\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow \left(1-\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}=\mathrm{dt}\\ \therefore \mathrm{I}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-1^2}=\frac{1}{2\times 1}\log \frac{\mathrm{t}-1}{\mathrm{t}+1}+\mathrm{C}\\ =\frac{1}{2}\log \frac{{\mathrm{x}}^2+1-\mathrm{x}}{{\mathrm{x}}^2+1+\mathrm{x}}+\mathrm{C}\end{array}$