$\int\left(\frac{x-6}{x^2-4}\right)dx$

$x^2-3x-1$

$12x^2\cdot\frac{x+1}{4x}$

$-16x^2+80x+96=0$

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

$\frac{-35\:-\:15}{-25}$

$\frac{x^2}{x^2+3}$