$\int\left(x+1\right)^2in\left(x+1\right)dx$
$-1-6-4+6-2-7-3-18$
$\frac{du}{dx}=\left(\frac{-u^2+u+2}{x\left(u-1\right)}\right)$
$w^4+2w^2+9$
$\left(mn^2-4\right)^2$
$\frac{18}{m}=-3$
$9a+3a-5a-7a$
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