$\int\frac{\left(\sqrt{x}+6\right)^{3}}{2\sqrt{x}}dx$
$10-9+2+4$
$v^2+18+324$
$\frac{d^{17}y}{dx^{17}}y=\left(-4cos\left(x\right)\right)$
$\int_{-1}^1\left(\sqrt{\left(4-x^2\right)}\right)dx$
$x^2+6x+10+7$
$x^2-3x=8$
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