$\int\frac{4x}{\sqrt{x-3}}dx$
$y'\:+\:5x^4y\:=\:x^4$
$3r^3\left(2r^3\right)$
$-2+4-4-3+5+4-5$
$\frac{d}{dx}8x\sqrt{16+0.5x}$
$\frac{\left(x^2y^4\right)^{-1}}{\left(x^4y^3\right)^{-2}}$
$a^3b^3x^3+1$
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