0\\\\\frac{1}{x+ \frac{1}{3} } >0\\\\x\in(- \frac{1}{3};+\infty) " alt="f(x)=ln(3x+1)\\\\f`(x)= \frac{(3x+1)`}{3x+1}= \frac{3}{3(x+ \frac{1}{3} )}= \frac{1}{x+ \frac{1}{3} }\\\\ f`(x)>0\\\\\frac{1}{x+ \frac{1}{3} } >0\\\\x\in(- \frac{1}{3};+\infty) " align="absmiddle" class="latex-formula">