$y\:=\:\left(4x^3\:+\:3x^2\right)^{tan\:x}$
$\int\:x\:\frac{3}{2}\:lnxdx$
$1-\left(-12\right)$
$6x^2+18x+24$
$-8n-4+n+7$
$1+\frac{3}{x+4}=\frac{3}{x^2+9x+20}$
$4u^2-36u+81$
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