$\tan\left(x\right)\left(1+\sec\left(2x\right)\right)$
$\frac{1}{y^2}dy=\frac{1}{x}dx$
$\left|-37\right|$
$7x-41+4y-2z=0$
$\left(x^2+4\right)\left(x^4+16\right)$
$\int\frac{2x^3-3x^2+3x-5}{\left(x^2+1\right)}dx$
$-3x+8y+4x-19-10y+z$
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