$\lim_{x\to\infty}\:0^{2x+1}$
$1x^2-4x+0$
$\int\frac{x}{\sqrt{1-x^2}}ln89dx$
$7x-8,\:x=4$
$\int\frac{7x}{\sqrt{\left(10x^2-5\right)}}dx$
$\left(5x^2+3x+4\right)-\left(2x^2+1\right)-\left(7x+3\right)$
$\frac{\left(z^4-3z^3+27^2-4z+4\right)}{\left(z-1\right)^{-1}}$
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