$\left(8n+4\right):2$
$64b^2-40b+100$
$1-20cd+100c^2d^2$
$\int\frac{\left(3arctan\left(x\right)-3x+x^3\right)}{x^4}dx$
$+\frac{2}{3}x+\frac{5}{6}x-\frac{7}{12}x$
$f\left(x\right)=\frac{x^2}{\left(1-x^3\right)^2}$
$23=x+11$
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