Question

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

Answer 1

$$\begin{gathered} \int \frac{e^{x}dx}{\sqrt{16-{\left(e^x\right)}^2}} \\ \mathrm{let}{e}^x=t \\ \Rightarrow e^{x}dx=dt \\ \mathrm{Now},\int \frac{e^{x}dx}{\sqrt{16-{\left(e^x\right)}^2}} \\ =\int \frac{dt}{\sqrt{16-t^2}} \\ =\int \frac{dt}{\sqrt{4^2-t^2}} \\ ={\sin }^{-1}\left(\frac{t}{4}\right)+C \\ ={\sin }^{-1}\left(\frac{e^x}{4}\right)+C \end{gathered}$$