Question

The value of ${\int }_1^e\left\{\frac{\left[1+\log \left(x\right)\right]}{3x}\right\}dx=$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $e$
• E. $\frac{1}{e}$

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_1^e\left\{\frac{\left[1+\log \left(x\right)\right]}{3x}\right\}dx$

$$\begin{gathered} I={\int }_1^e\left\{\frac{\left[1+\log \left(x\right)\right]}{3x}\right\}dx \\ I=\frac{1}{3}{\int }_1^e\frac{1}{x}dx+\frac{1}{3}{\int }_1^e\frac{\log \left(x\right)}{x}dx \end{gathered}$$

Let $\log \left(x\right)=t\Rightarrow \frac{1}{x}·dx=dt$

when $$\begin{gathered} x=1,t=0 \\ x=e,t=1 \end{gathered}$$

$$\begin{gathered} I=\frac{1}{3}{\int }_0^1\left(1\right)dt+\frac{1}{3}{\int }_0^{1}tdt \\ I=\frac{1}{3}{\left|t\right|}_0^1+\frac{1}{3}{\left|\frac{t^2}{2}\right|}_0^1 \\ I=\frac{1}{3}\left|1-0\right|+\frac{1}{3}\left|\frac{1}{2}-0\right| \\ I=\frac{1}{3}+\frac{1}{6} \\ I=\frac{1}{2} \end{gathered}$$

Hence, option $B$ is the correct answer.