$x'\left(t\right)=3e^t-x$
$0=6x-\frac{48}{x^2}$
$\int\frac{1}{\left(x-4\right)^2\left(x-7\right)}dx$
$2x^3y-8x^2y^2-6xy^3$
$5^0-2^5-\sqrt[3]{1}$
$\lim_{x\to\infty}\left(\frac{xe^{2x}}{2}\right)$
$\int x^3\left(1+x^4\right)^6dx$
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