Answer
$$\ln x-\ln |2x+1|+C$$
Work Step by Step
Given $$\int \frac{d x}{x(2 x+1)}$$
Since
\begin{align*}
\frac{1}{x(2 x+1)}&=\frac{A}{x}+\frac{B}{2x+1}\\
&=\frac{A(2x+1)+Bx}{x(2x+1)}\\
1&=A(2x+1)+Bx
\end{align*}
Then
\begin{align*}
\text{at } x&=0\ \ \ \ \ A=1\\
\text{at } x&=\frac{-1}{2}\ \ \ \ \ B=-2
\end{align*}
Hence
\begin{align*}
\int \frac{1}{x(2 x+1)}dx&=\int \frac{1}{x}dx+\int \frac{-2}{2x+1}dx\\
&=\ln x-\ln |2x+1|+C
\end{align*}
