Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{x^2+x+1}{x^2-x+1}\right)dx \\ \mathrm{Now}, \\ {x}^2-x+1\stackrel{1}{\overline{)x^2+x+1}} \\ {x}^2-x+1 \\ \underline{-+-} \\ \underline{2x} \\ \mathrm{Therefore}, \\ \frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1} \\ \Rightarrow \int \left(\frac{x^2+x+1}{x^2-x+1}\right)dx=\int dx+\int \left(\frac{2x-1+1}{x^2-x+1}\right)dx \\ =\int dx+\int \left(\frac{2x-1}{x^2-x+1}\right)dx+\int \frac{dx}{x^2-x+1} \\ =\int dx+\int \frac{\left(2x-1\right)dx}{x^2-x+1}+\int \frac{dx}{x^2-x+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2+1} \\ =\int dx+\int \frac{\left(2x-1\right)dx}{x^2-x+1}+\int \frac{dx}{{\left(x-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =x+\log \left|x^2-x+1\right|+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C \end{gathered}$$