1708$Let $I=\int_{-1}^{0}\frac{1}{x^4-x^2+1}dx=\frac{1}{2}.\int_{-1}^{0}\frac{(x^2+1)-(x^2-1)}{x^4-x^2+1}dx$\\\\ $I=\frac{1}{2}.\underbrace{\int_{-1}^{0}\frac{(x^2+1)}{x^4-x^2+1}dx}_{I_{1}}-\frac{1}{2}.\underbrace{\int_{-1}^{0}\frac{(x^2-1)}{x^4-x^2+1}dx}_{I_{2}}...............................(1)$\\\\ $I_{1}=\int_{-1}^{0}\frac{(x^2+1)}{x^4-x^2+1}dx$\\\\ Divide $N_{r}$ and $D_{r}$ by $x^2$, We get....\\\\ $I_{1}=\int_{-1}^{0}\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}dx=\int_{-1}^{0}\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+1^2}dx$\\\\ put $(x-\frac{1}{x})=t$, Then $(1+\frac{1}{x^2})dx=dt$,\\\\ So $\boxed{I_{1}=\int_{-1}^{0}\frac{1}{t^2+1^2}=tan^{-1}(x-\frac{1}{x})\big|_{-1}^{0}=-\frac{\pi}{2}}$
$Similarly $I_{2}=\int_{-1}^{0}\frac{(x^2-1)}{x^4-x^2+1}dx$\\\\ Divide $N_{r}$ and $D_{r}$ by $x^2$, We get....\\\\ $I_{1}=\int_{-1}^{0}\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}dx=\int_{-1}^{0}\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-(\sqrt{3})^2}dx$\\\\ put $(x+\frac{1}{x})=u$, Then $(1-\frac{1}{x^2})dx=du$\\\\ $\boxed{I_{2}=\int_{-1}^{0}\frac{1}{u^2-(\sqrt{3})^2}du=\frac{1}{2\sqrt{3}}.ln\left | \frac{u-\sqrt{3}}{u+\sqrt{3}} \right |\big|_{-2}^{\infty}=-\frac{1}{2\sqrt{3}}.ln\left | \frac{2+\sqrt{3}}{2-\sqrt{3}} \right |}$\\\\ Put value of $I_{1}$ and $I_{2}$ in equation.....(1), We Get\\\\\\\\ So $\textcolor[rgb]{0.0,1.0,0.0}{\boxed{\boxed{\textcolor[rgb]{1.0,0.0,0.0}{I=-\frac{1}{2}.\frac{\pi}{2}+\frac{1}{2}.\frac{1}{2\sqrt{3}}.ln\left | \frac{2+\sqrt{3}}{2-\sqrt{3}} \right |}}}}$\\\\ \underline{\underline{Correct Me If There is any mistake}}.........
