$\int x\left(x^3-3x+8\right)dx$

$\int\left(1+\frac{1}{y}\right)^3\left(\frac{1}{y^2}\right)dy$

$4x+7<3$

$3p\:+\:6\:=\:18\:\cdot$

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

$5^4\left(3^7\cdot2^5\right)^3\left(3^{-14}\cdot4^{-2}\cdot5\right)$

$x-6<9$