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Question
\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]
Sum
Solution
\[\text{ We have, }\]
\[I = \int\frac{e^x - 1}{e^x + 1}dx\]
\[ = \int\frac{2 e^x - \left( e^x + 1 \right)}{e^x + 1}dx\]
\[ = \int\frac{2 e^x}{e^x + 1}dx - \int dx\]
\[\text{ Putting e}^x + 1 = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \therefore I = \int\frac{2}{t}dt - \int dx\]
\[ = 2 \text{ log } \left| t \right| - x + C\]
\[ = 2 \text{ log} \left| e^x + 1 \right| - x + C\]
\[ = 2 \text{ log }\left( e^x + 1 \right) - x + C\]
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