Question

The value of the integral ${\int }_1^e$ $\frac{1+\log x}{3x}dx$ is equal to

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

Answer 1

Answer: (B)

Let $I=\underset{1}{\overset{e}{\int }}\frac{1+\log x}{3x}dx$
$=\underset{1}{\overset{e}{\int }}\frac{1}{3x}dx+\frac{1}{3}\underset{1}{\overset{e}{\int }}\frac{\log x}{x}dx$
Put $\log x=t$
$\Rightarrow \frac{1}{x}dx=dt$
$\therefore I=\underset{1}{\overset{e}{\int }}\frac{1}{3x}dx+\frac{1}{3}{\int }_0^{1}tdt$
$={\left[\frac{1}{3}\log x\right]}^e+\frac{1}{3}{\left[\frac{t^2}{2}\right]}_0^1$
$=\frac{1}{3}[\log e-\log 1]+\frac{1}{6}\left[1^2-0\right]$
$=\frac{1}{3}[1-0]+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$