Question

$\int \frac{x^2+x+1}{{\left(x+1\right)}^2\left(x+2\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\left(x^2+x+1\right)}{{\left(x+1\right)}^2\left(x+2\right)}dx \\ \mathrm{Let}\frac{x^2+x+1}{{\left(x+1\right)}^2\left(x+2\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{{\left(x+1\right)}^2}+\frac{\mathrm{C}}{x+2} \\ \Rightarrow \frac{x^2+x+1}{{\left(x+1\right)}^2\left(x+2\right)}=\frac{\mathrm{A}\left(x+1\right)\left(x+2\right)+\mathrm{B}\left(x+2\right)+\mathrm{C}{\left(x+1\right)}^2}{{\left(x+1\right)}^2\left(x+2\right)} \\ \Rightarrow x^2+x+1=\mathrm{A}\left(x^2+x+2x+2\right)+\mathrm{B}x+2\mathrm{B}+\mathrm{C}\left(x^{\mathit{2}}+2x+1\right) \\ \Rightarrow x^2+x+1=\left(\mathrm{A}+\mathrm{C}\right)x^2+\left(3\mathrm{A}+\mathrm{B}+2\mathrm{C}\right)x+\left(2\mathrm{A}+2\mathrm{B}+\mathrm{C}\right) \\ \mathrm{Equating}\mathrm{coefficient}\mathrm{of}\mathrm{like}\mathrm{terms}. \\ \mathrm{A}+\mathrm{C}=1.....\left(1\right) \\ 3\mathrm{A}+\mathrm{B}+2\mathrm{C}=1.....\left(2\right) \\ 2\mathrm{A}+2\mathrm{B}+\mathrm{C}=1.....\left(3\right) \\ \mathrm{Solving}\mathrm{these}\mathrm{three}\mathrm{equations}\mathrm{we}\mathrm{get} \\ \mathrm{A}=-2 \\ \mathrm{B}=1 \\ \mathrm{C}=3 \\ \mathrm{Hence},\frac{x^2+x+1}{{\left(x+1\right)}^2\left(x+2\right)}=\frac{-2}{x+1}+\frac{1}{{\left(x+1\right)}^2}+\frac{3}{x+2} \\ \therefore I=-2\int \frac{dx}{x+1}+\int \frac{d}{{\left(x+1\right)}^2}+3\int \frac{dx}{x+2} \\ =-2\log \left|x+1\right|-\frac{1}{x+1}+3\log \left|x+2\right|+\mathrm{C} \end{gathered}$$