$\int_{-\infty}^{-1}\left(\frac{1}{\left(x^2-6x\right)^{\frac{3}{2}}}\right)dx$
$\int x\left(4x^2-3\right)^3dx$
$4x^2+y^2+x+5y^2$
$\frac{d^3}{dx^3}\left(-x^2+3x-7\right)$
$\frac{dy}{dx}=\frac{10xy}{\left(lny\right)^{10}}$
$7x+8x-5+6x-7x+9x-12$
$x^2+3>\:4x$
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