Answer
\[I=\frac{3}{4}\ln |x-1|+\frac{1}{4}\ln |x+3|+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{x+2}{(x-1)(x+3)}dx\]
\[\frac{x+2}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}\;\;\;...(*)\]
\[x+2=A(x+3)+B(x-1)\]
\[x+2=(A+B)x+(3A-B)\]
Comparing coefficients both side
\[A+B=1\;\;\;...(1)\]
\[3A-B=2\;\;\;...(2)\]
Add equation (1) and (2)
\[4A=3\Rightarrow A=\frac{3}{4}\]
From (1)
\[B=\frac{1}{4}\]
From (*)
\[\frac{x+2}{(x-1)(x+3)}=\frac{3}{4(x-1)}+\frac{1}{4(x+3)}\]
\[\Rightarrow I=\int\left[\frac{3}{4(x-1)}+\frac{1}{4(x+3)}\right]dx\]
\[\Rightarrow I=\frac{3}{4}\ln |x-1|+\frac{1}{4}\ln |x+3|+C\]
Where $C$ is constant of integration
Hence ,
\[I=\frac{3}{4}\ln |x-1|+\frac{1}{4}\ln |x+3|+C\;.\]
