$x-3=\frac{3}{2}x+3$
$\int_0^x\left(\ln\left(1+t\right)\right)dt$
$\frac{dy}{dx}=\frac{y}{x}+xe^{\left(-3x\right)}$
$\lim_{x\to0}\left(\frac{ax^2+bx^3}{cx+dx^3}\right)$
$3x-x^3-4x+5+x^3+4x^2-6$
$\cot x+\tan x$
$m^2-4m+1=0$
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