$\sqrt{3}\cdot\:\:tgx-sin^2x=cos^2x$
$\frac{x^3}{x^2+2x^2}$
$dx-y^{2\:}dx+xydy-ydy=0$
$9x^2+36y^2+4z^2=36$
$y^2+3x=8$
$\lim_{x\to\infty}\left(\frac{1}{x}-\frac{4}{x^3}\right)$
$\int x\sqrt{2+\frac{1}{8}x^2}dx$
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