$\frac{x+1}{x+3}=2$
$\int\frac{4\left(x+1\right)}{\left(2x+1\right)}dx$
$xy+3x+y=6$
$\frac{dr}{dt}=r\left(1-r^2\right)$
$\left(x^6+x^3+1\right)\left(x^3-1\right)$
$\frac{d}{dx}\left(cos^3\left(tanx\right)\right)$
$x^2-16x+0$
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