$\int\frac{1}{64+x^2}dx$
$x5-5x^4+8x^3$
$x^2+8x-5=0$
$x2+6x+1=0$
$\frac{4x+2}{-x-1}=\frac{-2-4x}{x+1}$
$\lim_{x\to\infty}\left(\frac{\ln\left(5x+3\right)}{e^{\left(2x-4\right)}}\right)$
$9n^2-24n+16$
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