$\int_3^6\left(\frac{1}{\sqrt{36-x^2}}\right)dx$
$16x^2\left(y+x\right)^2$
$x^2\ge5$
$\int3edx$
$\frac{5^4}{5}$
$2x+7y-3x+y-x^2$
$16x^2+8x-3=0$
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