$x^3:x^3$
$\frac{-26}{9}+\frac{3}{4}=\frac{-65}{36}-\frac{1}{3}$
$\left(\frac{1}{5}\right)^{-4}$
$\int_1^{tan\:x}\left(\sqrt{6t+\sqrt{t}}\right)dt$
$-8\left(\:x\:-\:9\right)\:=\:32$
$\sqrt{x^4-x^2}$
$\int\left(\frac{6e^{-4x}}{e^{-4x}+7}\right)dx$
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