$\int\frac{x}{x^3-4x^2+4x}dx$
$\int\left(6\left(2x-3\right)^2\right)dx$
$\frac{x+2}{3}=\frac{2x-4}{2}$
$y=2-\frac{3x}{4+5x}$
$y^2-9y+8$
$\int\left(x^2\left(5x^3+5\right)^2\right)dx$
$y^2+y^2$
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