$y'=\frac{4x^2+y^2}{xy}$
$\frac{3x^4+4x^3+6x-2}{x-5}$
$\int\frac{7}{x^2-6x}dx$
$h\left(x\right)=\:\left(4x^2-3x^2+1\right)$
$3x+12a$
$-16-4\cdot4$
$\int x\left(6x-3\right)^4dx$
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