$\int_{-2}^2\left(\frac{1}{x^2+4}\right)dx$

$2x\sin\left(3y\right)dx+3x^2\cos\left(3y\right)dy=0$

$\int_{-2}^12xe^xdx$

$\frac{dy}{dt}=0.5t+y$

$\log\left(x-3\right)+\log\left(x+4\right)=\log\left(3x+12\right)$

$\left(6a-7b^b\right)\left(-3a+9b^2\right)$

$4\:+\:3\:x\:4\:-\:8\:x\:2$