$\int\left(\frac{\left(x^2-3x+8\right)}{\left(x^2-4x+7\right)^2}\right)dx$
$\left(2+4i\right)+\left(9+6i\right)$
$-4y^3-6y^3$
$-3-x-2x+3x^2$
$\left(2x-7\right)^2=\left(3x-1\right)^2$
$\int\frac{3x-2}{6x^2+x-2}dx$
$3n+3=15$
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