Answer
\[I=2\ln |x|-2\ln |x+1|-\frac{1}{x+1}+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{3x+2}{x(x+1)^2}dx\]
We first determine the partial fraction decomposition of the integrand
\[\frac{3x+2}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\;\;\;...(*)\]
\[3x+2=A(x+1)^2+Bx(x+1)+Cx\]
\[3x+2=A(x^2+2x+1)+Bx^2+Bx+Cx\]
\[3x+2=(A+B)x^2+(2A+B+C)x+A\]
Comparing like coefficients both side
\[A+B=0\Rightarrow B=-A\;\;\;...(1)\]
\[2A+B+C=3\;\;\;...(2)\]
\[A=2\]
From (1)
\[B=-2\]
From (2)
\[C=1\]
From (*)
\[\frac{3x+2}{x(x+1)^2}=\frac{2}{x}-\frac{2}{x+1}+\frac{1}{(x+1)^2}\]
\[\Rightarrow I=\int\left[\frac{2}{x}-\frac{2}{x+1}+\frac{1}{(x+1)^2}\right]dx\]
\[\Rightarrow I=2\ln |x|-2\ln |x+1|-\frac{1}{x+1}+C\]
Where $C$ is constant of integration
Hence ,
\[I=2\ln |x|-2\ln |x+1|-\frac{1}{x+1}+C\:.\]
