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Question
\[\int\frac{e^{2x}}{1 + e^x} dx\]
Sum
Solution
\[\int\frac{e^{2x} dx}{1 + e^x}\]
\[ \Rightarrow \int\frac{e^x . e^x}{1 + e^x}dx\]
\[\text{Let 1 }+ e^x = t \]
\[ \Rightarrow e^x = t - 1\]
\[ \Rightarrow e^x dx = dt\]
\[Now, \int\frac{e^x . e^x}{1 + e^x}dx\]
\[ = \int \frac{\left( t - 1 \right) . dt}{t}\]
\[ = \left( 1 - \frac{1}{t} \right)dt\]
\[ = t - \text{log }\left| t \right| + C\]
\[ = \left( 1 + e^x \right) - \log \left( 1 + e^x \right) + C\]
\[\text{Let C }+ 1 = C'\]
\[ = e^x - \text{log} \left( \text{1 + e}^x \right) + C'\]
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