$\lim_{n\to4}\left(\frac{4x^2-2x-8}{x-4}\right)$
$8x^2-6x\:-\:9$
$x^2+26x+99$
$1-4x>2$
$\int\frac{\cot\left(\ln x\right)}{x}dx$
$\frac{2+tan^2\left(u\right)}{sec^2u}-1$
$\int\sec\left(2x^2\right)\cdot\tan\left(2x^2\right)\left(4x\right)dx$
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