Question

\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]

Answer 1

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