$\int\left(\frac{1}{sqrt\left(4x-3-x^2\right)}\right)dx$
$y'\:=\:1\:+\:x\:+\:y\:+\:xy$
$6^4\cdot3^4$
$25x^2y^3-10x^3y^2$
$\frac{180\left(12-2\right)}{12}$
$-9+\left(-3\right)-\left(-2-1\right)-\left(-3+8\right)$
$\left(x^3+y^3\right)^4$
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