$\int\left(\left(2x+4\right)\left(x^2+4x+4\right)\right)dx$
$\frac { x ^ { 2 } - 16 } { x ^ { 2 } + 8 x + 16 }$
$12\:+\:3\:x\:4\:-\left(-2\right)$
$5\left(3m-6\right)$
$\frac{x^2+3x}{x^2+3}$
$-2b+4b-5b+12b-13b$
$12-6+8+3-5+9+7+8$
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