$\frac{tan\left(x\right)}{sec\left(x\right)-1}=\frac{\left(sec\left(x\right)+1\right)}{tan\left(x\right)}$
$8\cdot9+2\cdot5+3$
$\frac{dy}{dx}=y^2e^{-x}$
$x^2+16x+9$
$^{-3^2}$
$\frac{secx}{6tanx}$
$\int\left(-3x\left(-2x^2-1\right)^3\right)dx$
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