$y=\frac{2x+4}{1-x}$

$y^2=800-4x^2$

$3x^4-10x^2-8$

$\int\frac{\left(2-u\right)}{\left(u^2+1\right)}du$

$-5+-5+2+7+-35+65+-54$

$\frac{\left(\tan\:\left(x\right)-\sec\:\left(x\right)\right)}{1-\sin\:\left(x\right)}$

$\lim_{x\to3}\left(\frac{1-\cos\left(x\right)}{x^2}\right)$