$\frac{tan\left(6x\right)}{1-tan^2\left(6x\right)}$

$5\left(-5+2a\right)$

$\frac{1}{dx}=\frac{\sin\left(5x\right)}{dy}$

$\ln6+\ln\left(x-5\right)=\ln\left(x\right)$

$\left(x+2\right)\left(x-\frac{1}{2}\right)$

$\int\:\sqrt[3]{x^4}\left(2x^2+\frac{2}{x^4}-\frac{3}{\sqrt{3}}\right)dx$

$-\left(-1\right)^2+1^5$