$\ln\left(x^2+4x+4\right)$

$9y^{2}-y=0$

$1-\left(12^3\right)$

$\lim_{x\to0}\frac{cos\:\left(x\right)-e^x}{x^2+x}$

$-\left[1+2-3-4-\left(5-6+7\right)+8\right]-9\cdot\left(-10\right)-11$

$-cot\left(-x\right)sin\left(-x\right)tan\left(-x\right)$

$\frac{1}{2}\left(\cos\left(4x\right)-1\right)=\cos^2x$