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)

$\frac{1}{2}$

${\int }_1^e\frac{1+\log x}{3x}dx$

$I=\frac{1}{3}{\int }_1^e\frac{1}{x}dx+\frac{1}{3}{\int }_1^e\frac{\log x}{x}dx$

$=\frac{1}{3}[\ln x{]}_1^e+\frac{1}{3}\times \frac{1}{2}[(\ln x{)}^2{]}_1^e$

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{1}{2}$