$\int\frac{8}{x\left(1+x\right)^2}dx$

$\tan\left(d\right)+\frac{\cos\left(d\right)}{1+\sin\left(d\right)}=\sec\left(d\right)$

$-4\left(2\right)\left(27\right)$

$+\left(-1\right)^2-\left(-1\right)+1$

$\frac{x^2+4}{3}\frac{dy}{dx}=\frac{xy^2-x}{y}$

$\frac{2x^5-3x^3+x-4}{3x^4-x^2-3}$

$4+4a+4a^2$