1708\hspace{-16}\bf{\int\frac{x^2+1}{x^4+x^2+1}dx}$\\\\\\ Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{x^2}$\\\\\\ $\bf{\int\frac{1+\frac{1}{x^2}}{\left(x^2+\frac{1}{x^2}+1\right)}dx=\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{3}\right)^2}dx}$\\\\\\ Now Let $\bf{\left(x-\frac{1}{x}\right)=t\Leftrightarrow \left(1+\frac{1}{x^2}\right)dx=dt}$\\\\\\ $\bf{\int\frac{1}{t^2+\left(\sqrt{3}\right)^2}dt=\frac{1}{\sqrt{3}}.\tan^{-1}\left(\frac{t}{\sqrt{3}}\right)+\mathbb{C}}$\\\\\\ So $\bf{\int\frac{x^2+1}{x^4+x^2+1}dx = \frac{1}{\sqrt{3}}.\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+\mathbb{C}}$\\\\\\
