$x^2+4x-5\ge0$
$\frac{dy}{dx}=\frac{\sec^2x}{\tan^2y}$
$\int tan\:\left(9x-5\right)\:dx$
$\left(3xy^2-5yx\right)^2$
$\int\frac{1}{x\left(x-3\right)^2}dx$
$\int\frac{60x^4}{4x^5-5}dx$
$x^2+14x+40$
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