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प्रश्न
\[\int\frac{dx}{e^x + e^{- x}}\]
बेरीज
उत्तर
\[\int\frac{dx}{e^x + e^{- x}}\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[\text{ let } e^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[Now, \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{dt}{1 + t^2}\]
\[ = \tan^{- 1} \left( t \right) + c\]
\[ = \tan^{- 1} \left( e^x \right) + c\]
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