Question

$\int \frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\left(3x-2\right)dx}{{\left(x+1\right)}^2\left(x+3\right)} \\ \mathrm{Let}\frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{{\left(x+1\right)}^2}+\frac{C}{x+3} \\ \Rightarrow \frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}=\frac{A\left(x+1\right)\left(x+3\right)+B\left(x+3\right)+C{\left(x+1\right)}^2}{{\left(x+1\right)}^2\left(x+3\right)} \\ \Rightarrow 3x-2=A\left(x+1\right)\left(x+3\right)+B\left(x+3\right)+C{\left(x+1\right)}^2 \\ \Rightarrow 3x-2=A\left(x^2+x+3x+3\right)+B\left(x+3\right)+C\left(x^2+2x+1\right) \\ \Rightarrow 3x-2=\left(A+C\right)x^2+x\left(4A+B+2C\right)+3A+3B+C \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+C=0.....\left(1\right) \\ 4A+B+2C=3.....\left(2\right) \\ 3A+3B+C=-2.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right)\mathrm{we}\mathrm{get} \\ A=\frac{11}{4},B=-\frac{5}{2}\mathrm{and}C=-\frac{11}{4} \\ \therefore \frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}=\frac{11}{4\left(x+1\right)}-\frac{5}{2{\left(x+1\right)}^2}-\frac{11}{4\left(x+3\right)} \\ \Rightarrow I=\frac{11}{4}\int \frac{dx}{x+1}-\frac{5}{2}\int {\left(x+1\right)}^{-2}dx-\frac{11}{4}\int \frac{dx}{x+3} \\ =\frac{11}{4}\log \left|x+1\right|-\frac{5}{2}\left[\frac{{\left(x+1\right)}^{-1}}{-1}\right]-\frac{11}{4}\log \left|x+3\right|+C \\ =\frac{11}{4}\log \left|\frac{x+1}{x+3}\right|+\frac{5}{2\left(x+1\right)}+C \end{gathered}$$