$3+x=16$
$xy'-\frac{y}{x\left(1-x^2\right)}=0$
$\int\frac{\left(4x^3-t^2+16t\right)}{t^2+4}dx$
$-2,36-2,5$
$\int\left(6xcos\left(3x^2-2\right)\right)dx$
$\lim_{x\to0}\left(\frac{\sqrt{x-a}-\sqrt{a}}{x}\right)$
$\left(11x^{-3}y^2\right)\left(-3xy^{-3}\right)$
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