$ \int_{0}^{1} \frac{dx}{e^x + e^{-x}} $ is equal to
$ \frac{\pi}{4} - \tan^{-1}(e) $
$ \tan^{-1}(e) - \frac{\pi}{4} $
$ \tan^{-1}(e) + \frac{\pi}{4} $
$ \tan^{-1}(e) $
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