$\int\frac{x-3}{x\left(x^2-2x+3\right)}dx$
$3x^2+33x+72=0$
$\frac{4x^2-8x+18}{2x-4}$
$348+20$
$8x^4-24x^2+18$
$\frac{du}{dv}=\frac{\left(3u-2v\right)^2}{\left(2u-3v\right)^2}$
$23,5\:\cdot9,1$
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