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

$40x^2+8x+25$

$9x^2y^4-16z^2$

$\frac{\left(86\right)^{-6}}{\left(2b\right)^{-3}}$

$\lim_{x\to\infty}\left(-\frac{1}{tan\left(x\right)}\right)$

$\int\frac{3x^2-1}{\left(x+3\right)\left(x^2+4\right)}dx$

$\lim_{x\to0}\left(xln\left(1-x\right)\right)$