$\frac{24x^3y^2}{18xy^3}$

$\tan^2x+21=\sec^2x+20$

$2-3\cdot4-12$

$\left(2x-1\right)dx-dy=ydy$

$f\left(x\right)=\frac{-2}{\left(x-1\right)^2}+3$

$1^2+1^2\ln\left(1\right)$

$\int12\cdot\:\left(\frac{1}{12}+\frac{1}{3}x+\frac{1}{4}x^2\right)in\left(x\left(1+x\right)^2\right)dx$