$\int\frac{\sqrt{x^2-16\:}}{x^4}dx$
$\frac{2}{3}x+3$
$\left(+5\right)+\left(-1280\right)\left(-8\right)$
$-45-23-48$
$y'+\left(x+4\right)y=0$
$5+3\sqrt{x}=8$
$\csc^2\left(\frac{\pi\:}{11}\right)-\cot^2\left(\frac{\pi\:}{11}\right)$
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