Question

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

Answer 1

\[\int\frac{1}{e^x + 1}dx . . . (1)\]

Multiplying numerator and Denominator of eq (1) by e^x

\[\Rightarrow \int\frac{e^x \cdot dx}{e^x \left( e^x + 1 \right)}\]
\[\text{ Putting e}^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \Rightarrow \int\frac{dt}{t \left( t + 1 \right)}\]
\[ \therefore \frac{1}{t \left( t + 1 \right)} = \frac{A}{t} + \frac{B}{t + 1}\]
\[\frac{1}{t \left( t + 1 \right)} = \frac{A \left( t + 1 \right) + B t}{t \left( t + 1 \right)} . . . (2)\]
\[ \Rightarrow 1 = A \left( t + 1 \right) + B t\]
\[\text{ Putting  t } + 1\ = 0\text{ or}\, t\ = - 1\text{ in  eq  (2)  we  get}  , \]
\[ \Rightarrow 1 = A \times 0 + B \left( - 1 \right)\]
\[ \Rightarrow B = - 1\]
\[\text{ Now , putting  t = 0  in  eq  (2) we get , } \]
\[ \Rightarrow 1 = A \left( 0 + 1 \right) + B \times 0\]
\[ \Rightarrow A = 1\]
\[\text{ Putting  the  values  of  A  and  B  in eq (2) we  get } , \]
\[\frac{1}{t \left( t + 1 \right)} = \frac{1}{t} - \frac{1}{t + 1}\]
\[ \therefore \int\frac{dt}{t \left( t + 1 \right)} = \int\frac{dt}{t} - \int\frac{dt}{t + 1}\]
\[ = \text{ ln }\left| t \right| - \text{ ln }\left| t + 1 \right| + C\]
\[ = \text{ ln }\left| \frac{t}{t + 1} \right| + C\]
\[ = \text{ ln }\left| \frac{e^x}{e^x + 1} \right| + C\]
\[ = \text{ ln }e^x - \text{ ln }\left| e^x + 1 \right| + C\]
\[ = x - \text{ ln} \left| e^x + 1 \right| + C\]