$\int\frac{1+z^2}{z-z^3}dz$
$180+97+-3000$
$y\:=\:\frac{x\left(1+x^2\right)^3}{\left(1+x^3\right)^{\frac{1}{3}}}$
$x^2-3x-28\ge0$
$\int\left(x^2-2x+1\right)e^xdx$
$\lim_{x\to\infty}\left(\frac{lnx^4}{ln\left(x+1\right)^4}\right)$
$\cos^2x+\cos x=\sin^2x$
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