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प्रश्न
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उत्तर
\[\text{Let I} = \int\frac{1}{e^x + 1}dx\]
\[ = \int\frac{e^{- x}}{1 + e^{- x}}dx\]
\[Putting\ e^{- x} = t\]
\[ \Rightarrow - e^{- x} = \frac{dt}{dx}\]
\[ \Rightarrow e^{- x} dx = - dt\]
\[ \therefore I = \int\frac{- 1}{1 + t}dt\]
\[ = - \text{ln} \left| 1 + t \right| + C\]
\[ = - \text{ln } \left| 1 + e^{- x} \right| + C\]
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