$\left(\frac{1}{3}x+6\right)^3$
$\frac{dy}{dx}=\frac{\left(1-y\right)^4}{3}$
$\frac{3yx^{-3}}{xy^4}$
$x^2+6x=40$
$\lim_{x\to0}\left(-\arcsin\left(x\right)\cdot\cot\left(x\right)\right)$
$10^{\log\left(32\right)}=x$
$\int_1^{\infty}\left(\frac{1}{\sqrt{u}}\right)du$
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