$\lim_{x\to-1}\left(2x^2+1\right)\left(1-2x\right)^2$

$\frac{-16^2}{n}$

$\int\frac{2x+7}{\left(x+1\right)^2+4}dx$

$\frac{x^3-6x^2+2+5x^5}{x-2}$

$\left(2x-5\right)+\left(4x-2\right)$

$\int_1^{\infty}\left(\frac{1}{-1+x+x^2}\right)dx$

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