$\int\frac{\left(x^2+3x-4\right)}{\left(x^2-2x-9\right)}dx$

$16x^{2-40x+25}$

$2\:-2^2\:\:-1\:3\:-\:3-2\:3\:-\:\left[-5-1\:3\:-\:\left(3\:-1^3\:\:+\:2\:-4\:-1^2\:\right)\right]\:$

$\frac{d}{dx}\left(x^3+5xy+y^2\right)=x$

$4x+3y=18$

$\frac{d}{dx}x^2-xy=y^2$

$\int_{-1}^0\left(x\left(\sqrt{5+\left(3x^2+6\right)^2}\right)\right)dx$