$3\:+\:8g^2\:-\:5g^2\:+\:4g$
$\frac{d}{dt}v=\frac{1}{3}rh$
$\int\frac{y}{\sqrt{4-16\cdot y^2}}dy$
$\int _ { 2 } ^ { 2 } d x$
$\frac{dy}{dt}=\frac{y+3}{1-y}$
$-5\cdot\left(-2+6\right)+2$
$\int_0^v\left(\frac{1}{30-0.2v}\right)dv$
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