$\left(4\right)^5+\left(4\right)^3$

$\left(4x-2\right)\cdot\left(x+1\right)$

$\frac{d}{dx}ln\left[tg\left(\sqrt{x}\right)\right]^3\:+arc\:sec\left(y\right)$

$\lim_{x\to\infty}\left(\frac{3x^2+9}{4x^3+8}\right)$

$\left(-x^4-15x^3+9x^2-6x-8\right)+\left(7x^4+6x^32x^2-6x+1\right)$

$5x\:+12,con\:x=2,5$

$\sin^2\left(x\right)+cos^2\left(x\right)=\frac{1-\cos\left(4x\right)}{8}$