$-\frac{48c^2d^4}{-8cd}$

$\int\frac{t^3}{\sqrt{16-t^2}}dt$

$\int_{-1}^0\left(-\sqrt[2]{1-x^2}\right)dx$

$\left(2xy+2x+y+2\right)dx+\left(x^2+x-1\right)dy=0$

$y'=\left(y+x\right)^2$

$\left(\left(-5\right)+\left(-19\right)\right)-\left(\left(25\right)-\left(-23\right)\right)$

$2+3\:x\:2$