$\int\frac{1}{9\:}\:cot\left(3x\right)\:dx$
$3x\left(x+4\right)-2=2\left(x+8\right)$
$\left(4d^2+3\right)\left(4d^2-3\right)$
$\frac{\left(3x^3+2x^2-5x\right)}{\left(2x^5-4x^3+6x^2-8x+10\right)}$
$\frac{x}{-2}-5\le\:-10$
$\int\frac{x^3+1}{x-2}dx$
$\frac{dy}{dx}=\frac{2y-2x}{3y-2x}$
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