$\int\left(3+x^2\right)\sqrt{5+4x}dx$

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

$3\frac{dy}{dx}+12=4$

$156\:-\left(-45\right)$

$\frac{\left(1+tan\left(x\right)\left[tan\left(x\right)+sec\left(x\right)\right]\right)}{sec\left(x\right)+tan\left(x\right)}\:=\:sec\left(x\right)$

$\int2\cos\left(x\right)\cdot\cot\left(x\right)dx$

$x^2-6x-41=0$