$x^2-4x=6$
$\int\sin^2\left(3x\right)\cos^4\left(3x\right)dx$
$x\frac{dy}{dx}+4y=5x-3x^2$
$\frac{d}{dx}\left(4\sqrt{y}=x-3y\right)$
$\frac{1}{9x+18}$
$\frac{dy}{dx}=\frac{1}{ln\left(2x+y+3\right)}-2$
$\int\left(\frac{\left(7x\right)}{\left(7x^2+4\right)}\right)dx$
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