Question

$\underset{0}{\overset{\infty }{\int }}\frac{1}{1+e^x}dx$ equals

(a) log 2 − 1
(b) log 2
(c) log 4 − 1
(d) − log 2

Answer 1

(b) log 2

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\infty }\frac{1}{1+e^x}\mathit{d}x \\ \mathrm{Putting}{e}^x=t \\ \Rightarrow e^{x}dx=dt \\ \Rightarrow dx=\frac{dt}{t} \\ \mathrm{When}x\to 0;t\to 1 \\ \mathrm{and}x\to \infty ;t\to \infty \\ \therefore I={\int }_1^{\infty }\frac{1}{t\left(1+t\right)}\mathit{d}t \\ \mathit{=}{\int }_1^{\infty }\frac{\mathit{1}}{t\mathit{+}{t}^{\mathit{2}}}\mathit{d}t \\ ={\int }_1^{\infty }\frac{\mathit{1}}{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2}\mathit{d}t \end{gathered}$$

$$\begin{gathered} =\frac{1}{2\times {\displaystyle \frac{1}{2}}}{\left[\log \left|\frac{t+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}}{t+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right|\right]}_1^{\infty } \\ ={\left[\log \left|\frac{t}{t+1}\right|\right]}_1^{\infty } \\ ={\left[\log \left|\frac{{\displaystyle \frac{t}{t}}}{{\displaystyle \frac{t}{t}+\frac{1}{t}}}\right|\right]}_1^{\infty } \\ ={\left[\log \left|\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{1}{t}}}\right|\right]}_1^{\infty } \\ =\log \frac{{\displaystyle 1}}{{\displaystyle 1+0}}-\log \frac{1}{1+1} \\ =\log \left(1\right)-\log \left(\frac{1}{2}\right) \\ =0-\left(-\log 2\right) \\ =\log 2 \end{gathered}$$