If $\int \frac{f\left(x\right)}{x^2-x+1}\mathrm{d}x=\frac{3}{2}\log \left(x^2-x+1\right)+\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{2x-1}{\sqrt{3}}+C$  

then $f\left(x\right)$ is equal to 

Differentiating both the sides w.r.t. x , we get 

$\frac{f\left(x\right)}{x^2-x+1}=\frac{3}{2}\frac{2x-1}{x^2-x+1}+\frac{1}{\sqrt{3}}\frac{1}{1+({\displaystyle \frac{2x-1}{\sqrt{3}}}{)}^2}\left(\frac{2}{\sqrt{3}}\right)$

$\begin{array}{l}=\frac{3}{2}\frac{2x-1}{x^2-x+1}+\frac{1}{2\left(x^2-x+1\right)}\\ \Rightarrow f\left(x\right)=3x-1\end{array}$