Let \(I = \int_{0}^{1}\frac{dx}{e^{x}+e^{-x}}\)

Multiply numerator and denominator by \(e^x\):

\(I = \int_{0}^{1}\frac{e^{x}}{e^{2x}+1}dx\)

Let \(u = e^x\), then \(du = e^x dx\)

When \(x = 0\), \(u = e^0 = 1\)

When \(x = 1\), \(u = e^1 = e\)

So, \(I = \int_{1}^{e}\frac{du}{u^{2}+1}\)

The integral of \(\frac{1}{u^2 + 1}\) is \(\tan^{-1}(u)\)

Therefore, \(I = [\tan^{-1}(u)]_{1}^{e}\)

\(I = \tan^{-1}(e) - \tan^{-1}(1)\)

Since \(\tan^{-1}(1) = \frac{\pi}{4}\)

\(I = \tan^{-1}(e) - \frac{\pi}{4}\)