$\int\left(\left(x-3\right)\left(6x+4\right)\right)dx$
$-11t-5=-13-11t$
$9-\left(-5\right)-\left(-3-7-24\right)$
$3\left(8\right)^4-3\left(8\right)^3-\left(8\right)^2+3\left(8\right)-1\:$
$100\cdot0.01$
$\int x\cdot e^{40x}dx$
$\lim_{x\to\infty}\frac{x^4-10}{4x^3+x}$
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