$y'\:+\frac{1}{1+x^2}=\frac{1}{1+x^2}$
$\cos\left(x\right)-\sin\left(x\right)=\cos\left(x\right)\cot\left(x\right)$
$x^2+12+4$
$\sqrt{11^2}$
$4\left(5+2\left(5\cdot6-27\right)\right)$
$\left(6x^4-1\right)^3$
$x^5\cdot y^{-2}$
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