$45s^2+18s+1=0$
$\arctan\left(x\right).dx\:-\:tan\left(t\right).dt\:=0$
$\left(x+\frac{2}{5}\right)\left(x+3\right)$
$\left(x^2+1\right)\frac{dy}{dx}=x-xy$
$y=-4\left(3x^2+3x\right)^2-4\left(3x^2+3x\right)$
$-2.04^2$
$-\int\frac{y}{1+y^2}dy$
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