The given integral is $I = \int_{0}^{1}\frac{dx}{e^{x}+e^{-x}}$. Multiply the numerator and denominator by $e^{x}$ to rewrite the expression: $$I = \int_{0}^{1}\frac{e^{x}}{e^{2x}+1} dx$$
Let $u = e^{x}$. Then $du = e^{x} dx$. When $x = 0$, $u = e^{0} = 1$. When $x = 1$, $u = e^{1} = e$. The integral becomes: $$I = \int_{1}^{e}\frac{du}{u^{2}+1}$$
The integral of $\frac{1}{u^{2}+1}$ is $\tan^{-1}(u)$. Applying the limits: $$I = [\tan^{-1}(u)]_{1}^{e} = \tan^{-1}(e) - \tan^{-1}(1)$$ Since $\tan^{-1}(1) = \frac{\pi}{4}$, we have: $$I = \tan^{-1}(e) - \frac{\pi}{4}$$
Equating our result to the given expression $\tan^{-1}(e) + k$: $$\tan^{-1}(e) - \frac{\pi}{4} = \tan^{-1}(e) + k$$ Therefore, $k = -\frac{\pi}{4}$.
Final Answer: $-\frac{\pi}{4}$
AI generated content. Review strictly for academic accuracy.