$\frac{dy}{dx}+\frac{1}{x}y=x^3sen\left(2x\right)$
$y'=x^3y^3$
$12abc^3-3ba^3$
$3\frac{dy}{dx}=4y$
$\int\frac{8}{4+25\cdot w^2}dx$
$\left(2x-3\right)\left(3x+5\right)\left(6-4x\right)$
$\int\frac{\left(x^4+8x^2+8\right)}{\left(x^3-4x\right)}dx$
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