$\left(\frac{1+x^2}{3+4x^3}\right)^3$
$\int\left(\frac{1}{3+\sqrt{x+2}}\right)dx$
$-12+6\left(2.5x+6\right)$
$\left(\sqrt{x}\right)+\frac{1}{\sqrt{y}}=1$
$\lim_{x\to0\:}\left(1-\tan\left(2x\right)^{\frac{4}{x}}\right)$
$10+2u=18$
$\left(11a^2-6b\right)^2$
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