$\frac{x}{x^2-9}-\frac{1}{x-3}$
$\int-24x^{-8}dx$
$x\ge3+x$
$\frac{d}{dx}b^x$
$f\left(x\right)=\frac{1}{4}\left(x^5-9x^3+4\right)\left(5x-\frac{6}{x}\right)$
$\int\frac{3}{x^4-x^2+2}dx$
$\frac{x^3-9x^2+4+2}{x+3}$
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