Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\left(x^2+x-1\right)dx}{{\left(x+1\right)}^2\left(x+2\right)} \\ \mathrm{Let}\frac{x^2+x-1}{{\left(x+1\right)}^2\left(x+2\right)}=\frac{A}{x+1}+\frac{B}{{\left(x+1\right)}^2}+\frac{C}{x+2} \\ \Rightarrow \frac{x^2+x-1}{{\left(x+1\right)}^2\left(x+1\right)}=\frac{A\left(x+1\right)\left(x+2\right)+B\left(x+2\right)+C{\left(x+1\right)}^2}{{\left(x+1\right)}^2\left(x+2\right)} \\ \Rightarrow x^2+x-1=A\left(x^2+3x+2\right)+B\left(x+2\right)+C\left(x^2+2x+1\right) \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+C=1.....\left(1\right) \\ 3A+B+2C=1.....\left(2\right) \\ 2A+2B+C=-1.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ A=0 \\ B=-1 \\ C=1 \\ \therefore \frac{x^2+x-1}{{\left(x+1\right)}^2\left(x+2\right)}=\frac{-1}{{\left(x+1\right)}^2}+\frac{1}{x+2} \\ \Rightarrow I=\int \frac{-dx}{{\left(x+1\right)}^2}+\int \frac{dx}{x+2} \\ =-\int {\left(x+1\right)}^{-2}dx+\int \frac{dx}{x+2} \\ =-\left[\frac{{\left(x+1\right)}^{-2+1}}{-2+1}\right]+\log \left|x+2\right|+C \\ =\frac{1}{\left(x+1\right)}+\log \left|x+2\right|+C \end{gathered}$$