$3\frac{3}{14}\left(-4\frac{8}{9}\right)\left(7\right)\left(\frac{7}{8}\right)$
$\lim_{x\to0}\left(\frac{\sin^{-1}x-\tan^{-1}x}{x^3}\right)$
$f\left(x\right)\:=\:\frac{x}{x+7}\:+\:7\:$
$\int\frac{x^2+3}{x^4-1}dx$
$15-\left(-7\right)+1$
$\frac{x^2+1}{x^2-1}$
$5\left(3+3\right)$
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