$3\left(g\:+\:5\right)+2g$
$\frac{x+5}{x+1}+1=\frac{x-5}{x-2}$
$\int\frac{\left(6x-4\right)}{3x^2-4x}\:dx$
$xy+4y=x^{-3}e^x$
$1+2a-3a^2-6a^3$
$\lim_{x\to0}\left(\frac{6e^x-6}{x^3}\right)$
$\lim_{x\to infinity}\left(x^2\cdot\:e^{\left(x\right)}\right)$
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