$\int_{-\infty}^0\left(\frac{-36x}{\left(3x^2+1\right)^4}\right)dx$
$\sqrt{x^2+14x+49}=8$
$\left(2x-5y\right)\left(2x+8y\right)$
$\int\sec^2\left(4-6x\right)dx$
$\int\left(12x+5\right)^2dx$
$\left(4\:b\:+\:3\:\right)\:.\:\left(\:3b\:-\:10\right)$
$12,2-9$
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