(ln12)`=0
[m](lnu)`=\frac{1}{u}\cdot u`[/m]
[m](ln12+\frac{1}{3}ln(5x^3-2x))=(ln12)`+\frac{1}{3}(ln(5x^3-2x))`=0+\frac{1}{3}\cdot \frac{1}{5x^3-2x}\cdot (5x^3-2x)`=\frac{1}{3}\cdot \frac{1}{5x^3-2x}\cdot (15x^2-2)=\frac{15x^2-2}{3\cdot (5x^3-2x)}[/m]