$.3\int_0^{\infty}\left(e^{\left(-.3x\right)}\right)dx$
$a\left(x-2\right)+b\left(x-2\right)$
$3x^2-5x+4;\:x=6$
$\int\:\:t\:\left(t^2+1\right)^5dt$
$16x^2\left(-16x-24x\right)$
$x^2+6x=-11$
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