$x^3+3x^2+4x=0$
$\:\:\:\frac{1}{4}\int_0^4tdt$
$\int_1^{\infty}\left(\frac{e}{x}\right)dx$
$5x+1>-2x+3$
$-125+35+2$
$h\left(x\right)=x^3\left(zx\right)$
$\int x^2e^{1-4x}dx$
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