$\frac{1}{5}\left(6+3+1\right)^2$
$\int\frac{5x^2+20x+6}{\left(x\right)\left(x+1\right)^2}dx$
$\int\left(\frac{4}{4+y^2}\right)dy$
$x^3+5x^2-9x-45$
$a^4-4a^2-16$
$\int\:\:\left(y^3+5\right)^2dy$
$20x+40$
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