$1+tan^2\:sec^2$
$\int\frac{\left(x-3\right)^2}{\sqrt{1-x}}dx$
$\frac{dy}{dx}=\frac{2y}{x}+x^2e^x-1$
$x^2-3x-40\ge0$
$\int\frac{1+\sqrt{1-x^2}}{\sqrt{1-x^2}}dx$
$3x^2-x-1$
$y'\:=\:2\:+\frac{y^2}{x^2}$
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