$-10rt3r$
$\frac{dx}{dt}=x+8e^{2t}$
$\frac{3x+1}{x-4}\le1$
$\sec^2x-25=0$
$\left(1+tan^2\right)cosx=secx$
$\int\left(x^4\right)\left(\ln\left(x\right)\right)^2dx$
$\int_0^kx\left(x^2-32\right)dx$
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