$\int_{-1}^3\left(8x\right)dx$
$\left(0.6x-\frac{1}{2}y\right)^2$
$8-\left[-1-2-\left(-3+4\right)-2\right]$
$x^2-24x+25=0$
$\cos\left(3x\right)=\frac{-\sqrt{2}}{2}$
$\int\left(\tan\left(3u\right)\right)du$
$\lim_{x\to\infty}\left(\frac{7x^2+8x}{4x^3+7x+5}\right)$
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