Practise logarithmic standard forms

$\int(3x^2+\frac1x)\,dx=x^3+\ln|x|+C$, for $x\ne0$.

$\int\frac1{2-3x}\,dx=-\frac13\int\frac{-3}{2-3x}\,dx=-\frac13\ln|2-3x|+C$, for $x\ne2/3$.

$\int\frac{x}{1+x^2}\,dx=\frac12\int\frac{2x}{1+x^2}\,dx=\frac12\ln|1+x^2|+C$. This is the printed modulus form; since $1+x^2>0$, it also equals $\frac12\ln(1+x^2)+C$ for every real $x$.