$\int_{-1}^1\left(4x^3-6x^2+2x-1\right)dx$
$\sqrt[5]{m^5}$
$cos^3\left(x\right)=cos^3\left(x\right)-3sin^2\left(x\right)cos\left(x\right)$
$4x^2-15x+14$
$\frac{1+cos\left(y\right)}{cos\left(y\right)}=\frac{tan^2\left(y\right)}{sec\left(y\right)-1}$
$2y''\:-\:14y'\:+\:24y\:=\:0$
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