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Question
\[\int\frac{1}{e^x + e^{- x}} dx\]
Sum
Solution
\[\text{ Let I }= \int\frac{1}{e^x + e^{- x}}\text{ dx }\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{e^x dx}{\left( e^x \right)^2 + 1}\]
\[\text{ Putting e}^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \therefore I = \int\frac{dt}{t^2 + 1}\]
\[ = \tan^{- 1} t + C ..............\left( \because \int\frac{dt}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]
\[ = \tan^{- 1} e^x + C............. \left( \because t = e^x \right)\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{e^x dx}{\left( e^x \right)^2 + 1}\]
\[\text{ Putting e}^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \therefore I = \int\frac{dt}{t^2 + 1}\]
\[ = \tan^{- 1} t + C ..............\left( \because \int\frac{dt}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]
\[ = \tan^{- 1} e^x + C............. \left( \because t = e^x \right)\]
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