$\left(8\:x\:y\right)^2$

$\frac{d}{dx}\left(3x^5+4y^3=5-3y^5\right)$

$\left(y+3x\right)^2-\left(2x+2y\right)^3$

$\int_0^2\pi\left(x\right)dx$

$\frac{d}{dx}\left(\tan\left(y\right)+\ln\left(1+y^2\right)=\ln\left(x\right)\right)$

$\int\left(\frac{\left(3x^2-2x-7\right)}{x-2}\right)dx$