$x^4-1<0$
$( 5 x ^ { \frac { 1 } { 8 } } ) ^ { 2 }$
$\frac{tan^2t}{sec\:t}=sec\:t\:-\:cos\:t$
$\int\frac{10x-6}{\left(x^2-4\right)\left(x^2-2x+1\right)}dx$
$\int\left(\frac{1}{5-4x}\right)dx$
$\lim_{x\to3}\left(-6\cdotx^{6}\right)$
$m^2+17m+49$
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