$\int\left(r^2\cdot\sqrt{1+4r^2}\right)$
$\frac{dy}{dx}=\frac{x\left(y+3\right)^5}{3x^2-8}$
$5x+1;\:x=3$
$\frac{x}{4}-6=8$
$\left(-3\right)\cdot\left(+5\right):3$
$2\sin\:3x-1=0$
$-\:9t^2+5t-4t$
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