Question

Evaluate the following integrals:

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{{\left(x+1\right)}^2}+\frac{C}{x+3} \\ \Rightarrow 3x-2=A\left(x+1\right)\left(x+3\right)+B\left(x+3\right)+C{\left(x+1\right)}^2 \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2,x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 0=A+C\mathrm{and}3=4A+B+2C\mathrm{and}-2=3A+3B+C \\ \mathrm{or}A=\frac{11}{4}\mathrm{and}B=-\frac{5}{2}\mathrm{and}C=-\frac{11}{4} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{11}{4}}}{x+1}+\frac{{\displaystyle -\frac{5}{2}}}{{\left(x+1\right)}^2}+\frac{-{\displaystyle \frac{11}{4}}}{x+3}\right)\mathrm{d}x \\ =\frac{11}{4}\int \frac{1}{x+1}\mathrm{d}x-\frac{5}{2}\int \frac{1}{{\left(x+1\right)}^2}\mathrm{d}x-\frac{11}{4}\int \frac{1}{x+3}\mathrm{d}x \\ =\frac{11}{4}\log \left|x+1\right|+\frac{5}{2\left(x+1\right)}-\frac{11}{4}\log \left|x+3\right|+c \\ \mathrm{Hence},\int \frac{3x-2}{{\left(x+1\right)}^2\left(x+3\right)}\mathrm{d}x=\frac{11}{4}\log \left|x+1\right|+\frac{5}{2\left(x+1\right)}-\frac{11}{4}\log \left|x+3\right|+c \end{gathered}$$