$\int\left(3x^4-4x-8\right)\left(12x^3-4\right)dx$
$6x^2+6$
$\frac{1}{4}\int_0^4\left(x\right)dx$
$\lim_{x\to\infty}\left(x-3\right)^{\frac{4}{x}}$
$\left(-4\right)\left(-2\right)-6\left(2-5\right)$
$\left(\left(1+2x\right)^x=y\right)$
$\frac{3x^3-17x^2+27x-12}{x-4}$
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