$\int_{0}^{+\infty}\frac{\arctan^{2}\left(z\right)}{z^{4}}dz$
$x^4+x^2-\frac{8}{x-3}$
$x^2-2x+\frac{3}{x}-1$
$120\cdot3$
$27x^2+16x+6=0$
$-1-3\csc\left(x\right)=-7$
$\frac{2x^5+6x^3-x}{x^2}$
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