$$0^2\cdot2^2$$

$$\int\left(16sec^2\left(8x\right)tan\left(8x\right)\right)dx$$

$$\:-x^3+4x^2-5x+2$$

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

$$y'+\frac{y}{x}=xy^{\frac{1}{3}}$$

$$\lim_{x\to\infty}\left(\frac{1}{6}x^3\right)$$