$3x^2+x^2+x\:$
$\int\frac{\tan^2\left(x\right)}{sec^5\left(x\right)}dx$
$\int\frac{3x+12}{x^2-4}dx$
$\left(x+1\right)\frac{dy}{dx}-x=6$
$\frac{x^3-6x-10}{x+3}$
$\frac{8x^2+16x+6}{2x+3}$
$y^2+34y+1=0$
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