$-1+\frac{45}{15}$
$\int\:\left(4x^2-2x+3\right)^2\left(-4x-1\right)dx$
$\lim_{x\to\infty}\left(\frac{2x^3+16}{x^2+4}\right)$
$6x+1+7x$
$\int\left(x^3+x\right)\left(\sqrt{x^2+3}\right)dx$
$\frac{\left(x^2+y\right)^3}{x^2+y}$
$6x+3x\left(4x+2\right)$
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