$\sqrt{5}\int\left(\tan^2x\right)\left(\sec x\right)dx$
$3x-4y=6$
$1+2+4$
$\int\frac{x^2-2x}{x^3-3x^2}\:dx$
$1+\cos2x-\cos^2x=\cos^2x$
$\frac{dy}{dx}=\frac{4}{9}y$
$\frac{dy}{dx}=10y^2$
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