$x^2+4x+8$
$\frac{dy}{dx}=\frac{1}{2x+y}$
$\log\:_{17}\left(7-2x\right)=\log\:_{17}13$
$\int\frac{1}{\left(x-3\right)x\left(x+2\right)}dx$
$x^2=\frac{16}{169}$
$\frac{x^4-1}{1-x^2}$
$\int\frac{2x-3}{\sqrt{4-x^2}}dx$
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