$\:-96\:-\:\left(24\:-\:47\right)\:+\:\left(38\:-\:-31\right)\:-\:\left(-17\:-\:-54\right)$

$2x^2-7x+3$

$\frac{-3}{x}+\frac{3}{x^2}=0$

$-4-\frac{1}{4}\tan\left(x\right)=-\frac{17}{4}$

$\int\left(8x^5-10x^4+4x^2-8x-3\right)dx$

$\lim_{x\to\infty}\left(x^{\frac{4}{3}}\right)$

$x\frac{1}{4}\left(2x\frac{3}{4}-x\frac{1}{4}\right)$