$15a^2-8a^2+a$
$\frac{x^4-5x^2+4}{x+5}$
$-9\:x\:8$
$\frac{\tan\left(x\right)}{1+\sec\left(x\right)}=\frac{1+\sec\left(x\right)}{\tan\left(x\right)}$
$\int\frac{-x^2+4x}{-4}-\left(-x^2+4x\right)dx$
$\int x^23^{1-x^3}dx$
$x^2+4x+6=0$
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