Simplifying and decomposing rational integrands

First. $\frac{x+1}x=1+\frac1x$, so the integral is $x+\ln|x|+C$.

Second. $\frac x{x+1}=1-\frac1{x+1}$, so the integral is $x-\ln|x+1|+C$.

Third. We seek $\frac1{x(2-3x)}=\frac Ax+\frac B{2-3x}$, i.e. $1=A(2-3x)+Bx$. Taking $x=0$ gives $A=\frac12$; taking $x=\frac23$ gives $B\cdot\frac23=1$, so $B=\frac32$. Hence
$$\int\frac{dx}{x(2-3x)}=\frac12\ln|x|+\frac32\cdot\left(-\frac13\right)\ln|2-3x|+C=\frac12\ln\left|\frac x{2-3x}\right|+C.$$

Remark. Differentiate your answer to check: $\frac12\left(\frac1x+\frac3{2-3x}\right)=\frac12\cdot\frac{2}{x(2-3x)}$ ✓.