$\lim_{x\to0}\left(\frac{\frac{1}{\left(1+x\right)}-1}{x}\right)$
$\frac{8x^2+10x+3}{4x+3}$
$3tanx+3=0$
$\left(cotx\:-cscx\right)\left(cosx+1\right)$
$2025\left(+135\right)$
$\int\frac{2t^2+11t+8x}{t^3+4t^2+4t}dx$
$\frac{x^3-3x^2-4}{x-1}$
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