Question

The value of ${\int }_0^1\frac{1}{e^x+e}dx$ is

• A. $\left(\frac{1}{e}\right)\log \left[\left(\frac{1+e}{2}\right)\right]$
• B. $\log \left[\left(\frac{1+e}{2}\right)\right]$
• C. $\left(\frac{1}{e}\right)\log (1+e)$
• D. $\log \left[\frac{2}{\left(1+e\right)}\right]$
• E. $\left(\frac{1}{e}\right)\log \left[\frac{2}{\left(1+e\right)}\right]$

Answer 1

The correct option is A

$\left(\frac{1}{e}\right)\log \left[\left(\frac{1+e}{2}\right)\right]$

Explanation for the correct answer.

Evaluate the integral.

$$\begin{gathered} {\int }_0^1\frac{1}{e^x+e}dx \\ ={\int }_0^1\frac{1}{e^x\left(1+{\displaystyle \frac{e}{e^x}}\right)}dx \end{gathered}$$

Let,

$$\begin{gathered} 1+\frac{e}{e^x}=t \\ \Rightarrow -\frac{e}{e^x}dx=dt \\ \Rightarrow \frac{1}{e^x}dx=-\frac{1}{e}dt \end{gathered}$$

$$\begin{gathered} I=-\frac{1}{e}{\int }_{1+e}^2\frac{dt}{t} \\ =-\frac{1}{e}{\left[\log t\right]}_{1+e}^2 \\ =-\frac{1}{e}\left[\log \frac{2}{1+e}\right] \\ =\frac{1}{e}\left[\log \frac{1+e}{2}\right] \end{gathered}$$

Hence, option A is correct .