$u^2-6u+u$
$\cot\left(\sec-1\right)\left(\sec+1\right)$
$\lim_{x\to\infty}\left(\frac{6x^2+x+4}{2x^2-4x+5}\right)$
$\frac{dy}{dx}=\left(\sin\left(2x\right)+\cos\left(2x\right)\right)^2$
$5\sqrt[3]{128a}+\sqrt[3]{2a}-\sqrt[3]{16a}$
$x\:^2\:+2x=8$
$x\sqrt[5]{\left(x+1\right)}^2$
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