$36y^2+24y+4$
$\frac{\left(3x^4+5x^3+x^2+3\right)}{\left(x^2-3x+3\right)}$
$-\int\left(\frac{x}{\left(1+x^2\right)}\right)dx$
$y\cdot\:\:p'-p^2=0$
$\int\frac{3x+2}{\left(3x^2+4x\right)^4}dx$
$\frac{3}{x+11}+\frac{8}{x+11}$
$\tan^2x+21=\sec^2x+20$
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