$\int\frac{1}{\sqrt{16-49x^2}}dx$

$\left(-4\right)^4\cdot\left(\left(-4\right)^4\right)^3$

$f\left(x\right)=\:\left(-x^3+4\right)\ln\left(2x\right)$

$-16x-x$

$\tan\left(\frac{5cm}{12cm}\right)$

$\int\:\:3^{x+1}dx$

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