$1+\frac{1}{x-1}=\frac{6}{x^2-1}$
$\int\left(2x+3\right)\left(x^2+3x+4\right)^2dx$
$\left(2^3\right)^5$
$\lim_{x\to0}\left(\frac{\pi}{2}-x\right)\tan\left(x\right)$
$\frac{3x+1}{x-4}\le1$
$\left(9xy^3+11z^4\right)^2$
$\int\left(-\frac{1}{2}\right)dx$
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