$\int\left(\frac{x^4+3x^2+4}{\left(x+1\right)^2}\right)dx$
$\left(10x-2xy\right)^2$
$t^3-8t^2+t$
$3x^2=0$
$\left(\frac{7}{3}x-4\right)^2$
$2\cdot7-4$
$\frac{x+2}{3}+\frac{4x+1}{2}$
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