$\left(\frac{1}{4}v\right)^2\:=\:v^2$
$3^4+4^3-6^2$
$\lim_{x\to+\infty}\left(\frac{\sqrt{x^5+2x-6}}{x^3-4x+2}\right)$
$^{x^6-36}$
$\int\left(cos\left(x^4\right)\right)dx$
$2\left(5+8\right)-3\left(\:6\cdot3+2\right)\:$
$\frac{\frac{2}{2x-1}-1}{4x-2+1}$
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