$tan^2x+tanx=42$
$\frac{sin^2\left(-x\right)}{\tan\:^2\left(x\right)}=\cos\:^2\left(x\right)$
$\int\frac{3x+3}{x^2+4x+4}dx$
$2\cdot\left[3-\left(-8\right)\right]$
$y'-5y=-2x$
$\frac{4y^2-1}{4y^2+14y-8}$
$-5+6x\ge4x+7$
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