$\int\frac{x^2+x+10}{\left(2x-3\right)\left(x^2+4\right)}dx$
$-x+1+1$
$\left(+62\right)\cdot\left(+34\right)$
$\left(-7+2\right).3$
$12+\left(13-2.4\right).6$
$81\:a^4-\:100\:b^4$
$\lim_{x\to2}\left(\frac{5\left(x-2\right)\sqrt{x+7}}{3-\sqrt{x+7}}\right)$
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