$\int \frac{x^{2}+x+1}{x^{2}-x+1} d x$

ধরি, $\mathrm{x}^{2}+\mathrm{x}+1=\mathrm{A}\left(\mathrm{x}^{2}-\mathrm{x}+1\right)+\mathrm{B}(2 \mathrm{x}-1)+\mathrm{C}$ উভয় পক্ষ থেকে $x^{2}, x$ B $x^{0}$ এর সহগ সমীকৃত করে পাই,

$\begin{array}{l} A=1 ;-A+2 B=1 ; A-B+C=1 \\ \therefore B=1 \Rightarrow 1-1+C=1 \therefore C=1 \\ \therefore \int \frac{x^{2}+x+1}{x^{2}-x+1} d x=\int \frac{\left(x^{2}-x+1\right)+(2 x-1)+1}{x^{2}-x+1} d x \\ =\int\left(1+\frac{2 x-1}{x^{2}-x+1}+\frac{1}{x^{2}-x+1}\right) d x \end{array}$

$\begin{array}{l}=x+\ln \left(x^{2}-x+1\right)+\int \frac{d x}{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ =x+\ln \left(x^{2}-x+1\right)+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+C\end{array}$