Question

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals

Options

•  log 2 − 1

•  log 2

• log 4 − 1

•  − log 2

Answer 1

 log 2 

\[\text{We have}, \]
\[I = \int_0^\infty \frac{1}{1 + e^x} d x\]
\[\text{Putting } e^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \Rightarrow dx = \frac{dt}{t}\]
\[\text{When}\ x \to 0; t \to 1\]
\[\text{and }x \to \infty ; t \to \infty \]
\[ \therefore I = \int_1^\infty \frac{1}{t\left( 1 + t \right)} d t\]
\[ = \int_1^\infty \frac{1}{t + t^2} d t\]
\[ = \int_1^\infty \frac{1}{\left( t + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2} d t\]

\[= \frac{1}{2 \times \frac{1}{2}} \left[ \log\left| \frac{t + \frac{1}{2} - \frac{1}{2}}{t + \frac{1}{2} + \frac{1}{2}} \right| \right]_1^\infty \]

\[ = \left[ \log\left| \frac{t}{t + 1} \right| \right]_1^\infty \]

\[ = \left[ \log\left| \frac{\frac{t}{t}}{\frac{t}{t} + \frac{1}{t}} \right| \right]_1^\infty \]

\[ = \left[ \log\left| \frac{1}{1 + \frac{1}{t}} \right| \right]_1^\infty \]

\[ = \log\frac{1}{1 + 0} - \log\frac{1}{1 + 1}\]

\[ = \log\left( 1 \right) - \log\left( \frac{1}{2} \right)\]

\[ = 0 - \left( - \log2 \right)\]

\[ = \log2\]