$\left(2n+2\right)\left(2n+1\right)$
$y=\frac{\left(4x^2+1\right)}{\left(x+3\right)^2}$
$\tan^2x+5=\sec^2x+4$
$5x-4y-3x+2y$
$\lim_{x\to\infty}\left(1+\frac{12}{x}\right)^{\frac{x}{6}}$
$y=\sqrt{ax}+\frac{a}{\sqrt{ax}}$
$\ln\left[\frac{\left(2x+1\right)^3}{\left(x-4\right)}\right]$
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