$$\int7\cdot sin^2\left(x\right)\cdot\cos^3\left(x\right)dx$$

$$x^2-3x+4>0$$

$$9x-7x+9x-5$$

$$\int\left(14xe^{-7x}\right)dx$$

$$\frac{x\:+\:4}{x^2\:+\:x\:-\:12}$$

$$\frac { x ^ { 9 } } { x ^ { 4 } }$$

$$\sqrt{1+4x^2}dy=y^3xdx$$