$\int\:\frac{x^4}{\:\left(x^2-1\right)\left(x+2\right)}dx$
$\frac{d}{dx}\frac{1}{\left(4x^2+3\right)^4}$
$\lim_{x\to\infty}\left(7+\frac{1}{3x}\right)$
$\frac{3}{1}\cdot\frac{x^2}{2}$
$-3x+1+-3x+1$
$\left(-3\right).\left(-4\right).\left(-2\right).\left(+4\right).\left(-3\right)$
$2x+x+x$
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