$8\:x\:-6$
$\frac{1}{y^2}y'\:=8x^3$
$\frac{x^3+6x-10}{x+2}$
$\cos\left(x\right)^2=\frac{1+\cos\left(2x\right)}{2}$
$\left(6x+9y^2\right)$
$\int\frac{x^2}{\sqrt{25-x}}dx$
$\frac{dy}{dx}-\left(x+2\right)=0$
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