inverse tan integral

\hspace{-16}\mathbf{\int_{\frac{1}{2}}^{2}\frac{\tan^{-1}(x)}{x^2-x+1}dx}

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xYz ·

\texttt{Substituting x->1/x }..\\ I=\int_{\frac{1}{2}}^{2}{\frac{\cot^{-1} x}{x^2-x+1}} \\ 2I=\int_{\frac{1}{2}}^{2}{\frac{\tan^{-1} x+ \cot^{-1} x}{x^2-x+1}} \\ I=\frac{\pi}{4}\int_{\frac{1}{2}}^{2}{\frac{1}{x^2-x+1}} \\

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