$\frac{x^2}{91}-\frac{x}{18}+\frac{1}{16}$

$\lim_{x\to0}\frac{\tan^2\left(4x\right)}{1-\cos\left(5x\right)}$

$\left(-2x^3-2x^2+9x+1\right)\cos\left(x\right)+\left(6x^2+4x-9\right)\sin\left(x\right)+c$

$2\cdot3\cdot\left(3+4\right)$

$2\left(7x^3-8x+12\right)$

$9x^2-12x+36$

$\left(\frac{3}{2}-2y\right)^2$