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

$\left(8+x^2\right)^2$

$x^2+2y=1$

$\lim_{x\to0}\left(\left(\frac{\ln\left(x^2+x\right)}{2+\ln\left(x\right)}\right)\right)$

$\left(x-5\right)\left(x+1\right)\left(x-2\right)$

$0.02\cdot x^3-0.4\cdot x^2+10x+1000;\:x=100$

$-4y-4y+5y^2+3y-6y^2+8y+5y$