$\tan\left(x\right)+2\sin^2\left(x\right)$
$x^2-4x>-3$
$\frac{dy}{dx}=\frac{3y}{x}+x^5$
$125u^4-20n^4$
$\int\frac{\left(x^2+2x+1\right)}{\left(x-1\right)\left(x^2+1\right)^2}dx$
$\frac{d}{dx}x^3+x+y^3+3y=6$
$x^2+6x-16$
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