$\int\left(\frac{9sin^3\left(x\right)}{cos\left(x\right)}\right)dx$
$sen\:x\:tg\:x\:+1=sen\:x\:+tg\:x$
$4y'=y^3cosx$
$\frac{3x^3+6x^2-4x+3}{x-3}$
$\cos\left(6-x\right)$
$\frac{dy}{dx}=\frac{x}{y^2},\:y\left(0\right)=1$
$\int\frac{\left(6x-7\right)}{\left(x+1\right)^3}dx$
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