Question

$\int \frac{x^3+x^2+2x+1}{x^2-x+1}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{x^3+x^2+2x+1}{x^2-x+1}\right)dx \\ {x}^2-x+1\stackrel{x+2}{\overline{)x^3+x^2+2x+1}} \\ {x}^3-x^2+x \\ \underline{-+-} \\ 2x^2+x+1 \\ 2x^2-2x+2 \\ \underline{-+-} \\ 3x-1 \\ \mathrm{Therefore}, \\ \frac{x^3+x^2+2x+1}{x^2-x+1}=x+2+\frac{3x-1}{x^2-x+1}.....\left(1\right) \\ \mathrm{Let} \\ 3x-1=A\frac{d}{dx}\left(x^2-x+1\right)+B \\ 3x-1=A\left(2x-1\right)+B \\ 3x-1=\left(2A\right)x+B-A \\ \mathrm{Equating}\mathrm{Coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ 2A=3 \\ A=\frac{3}{2} \\ B-A=-1 \\ B-\frac{3}{2}=-1 \\ B=\frac{1}{2} \\ \int \left(\frac{x^3+x^2+2x+1}{x^2-x+1}\right)dx=\int \left(x+2\right)dx+\int \left(\frac{{\displaystyle \frac{3}{2}\left(2x-1\right)+\frac{1}{2}}}{x^2-x+1}\right)dx \\ =\int \left(x+2\right)dx+\frac{3}{2}\int \left(\frac{2x-1}{x^2-x+1}\right)dx+\frac{1}{2}\int \frac{dx}{x^2-x+1} \\ =\int \left(x+2\right)dx+\frac{3}{2}\int \frac{\left(2x-1\right)dx}{x^2-x+1}+\frac{1}{2}\int \frac{dx}{x^2-x+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+1} \\ =\int \left(x+2\right)dx+\frac{3}{2}\int \frac{\left(2x-1\right)dx}{x^2-x+1}+\frac{1}{2}\int \frac{dx}{{\left(x-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{x^2}{2}+2x+\frac{3}{2}\log \left|x^2-x+1\right|+\frac{1}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)+C \\ =\frac{x^2}{2}+2x+\frac{3}{2}\log \left|x^2-x+1\right|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C \end{gathered}$$