$\left(3x^2+x\right)\cdot\left(x+2\right)$

$y'=9.81-2\cdot y^2$

$\frac{y}{\sqrt{1+y^2}}dy=\frac{x}{\sqrt{x}^2}dx$

$\frac{1}{x^4+4x^2+4}$

$\left(-xy+8\right)\left(xy+8\right)$

$4x^2-12x+8$

$\frac{1-\cos\left(x\right)}{\tan\left(x\right)^2}=\frac{\cos\left(x\right)}{\sec\left(x\right)+1}$