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Let \(I = \int\frac{e^{x}}{\sqrt{4-e^{2x}}}dx\)
Substitute \(u = e^{x}\), then \(du = e^{x}dx\)
So, \(I = \int\frac{1}{\sqrt{4-u^{2}}}du\)
We know that \(\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx = \sin^{-1}(\frac{x}{a})+C\)
Here, \(a = 2\), so \(I = \sin^{-1}(\frac{u}{2})+C\)
Substituting back \(u = e^{x}\), we get \(I = \sin^{-1}(\frac{e^{x}}{2})+C\)
Correct Answer: \(\sin^{-1}(\frac{e^{x}}{2})+C\)
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