Evaluate the following integrals:

Let $\mathrm{e}^{x}=\mathrm{t}$

Then we have, $\mathrm{e}^{x} \mathrm{dx}=\mathrm{dt}$

Therefore, $\int \frac{e^{x}}{\sqrt{16-e^{2 x}}} d x=\int \frac{d t}{\sqrt{4^{2}-t^{2}}}$

Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$

Hence, $\int \frac{d t}{\sqrt{4^{2}-t^{2}}}=\sin ^{-1}\left(\frac{e^{x}}{a}\right)+c$