$\int\left(\frac{5}{\left(3x-6\right)^2}\right)dx$
$3x^8+11x^6-13x^5+9x^3-7^2+13$
$\frac{dy}{dx}=3x^5$
$2y''+12y'+18y=0$
$1x^2+2x-1$
$x\:+\:2\:\le\:6$
$\frac{du}{dr}=\frac{5+\sqrt{r}}{2+\sqrt{u}}$
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