$\frac{3}{x+1}=\frac{x}{x+1}$
$a^2+16a+64$
$\frac{x^3+6x+2}{x-3}$
$\frac{4x^2+21x+48}{x+7}$
$\int8e^{\frac{1}{4}t}\:\:sin\left(2t\right)dt$
$1x^2\:+\:18x\:+\:1c$
$\frac{1}{12}=b^4$
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