$-2\left(3x-2\right)=-2$
$\int\frac{2x^2}{\left(x-3\right)^2}dx$
$\frac{\left(2x+15\right)}{\left(5\cdot\left(x+3\right)\cdot7\right)}$
$y'=1-4y^2$
$6t^2+36=0$
$\tan\left(\frac{\sqrt{10}}{6}+\frac{\sqrt{5}}{4}\right)$
$xy'=y+x$
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