$\left(x+3\right)^2\:y\:\left(x-3\right)^2$

$x^2\:-\:\frac{3}{2}x\:+\:\frac{1}{5}$

$\int_0^{\frac{\pi}{7}}\left(3\:sec^4x\:tan^3x\right)dx$

$\lim_{x\to\infty}\left(\frac{ln\left(x^{12}\right)}{4x}\right)$

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

$64b^4-36a^2$

$x^2\:+16^2+\:64\:\:$