$\int_{-\infty}^6\left(\frac{6}{x^2+4}\right)dx$

$\left(x+3\right)^2=\left(2x+1\right)^2+8$

$\left(2m^3+5m^2n^3-8m^4\right)\left(-6mn^2\right)$

$\frac{dx}{dy}=e^{-y}$

$\frac{x^3-3x+2}{x-1}$

$\left(\left(x^2+3\right)\left(\left(4x-9\right)\right)\right)$

$2+\frac{2k}{n}$