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

$\frac{x^6}{\left(1+4x\right)^2}$

$\lim_{x\to\infty}\left(x-\left(x^2-8x-4\right)^{\frac{1}{2}}\right)$

$4x^2+4x+xy$

$-\left(-12\right)+\left(+11\right)+\left(-12\right)-0-\left(+11\right)$

$-2n+5+1$

$e^{2.9675}$