Definite Integration Question 404
Question: $ \int_1^{e}{\frac{1+\log x}{x}dx=} $
[SCRA 1986]
Options:
A) $ \frac{3}{2} $
B) $ \frac{1}{2} $
C) $ \frac{1}{e} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_1^{e}{\frac{1+\log x}{x}}dx=\int_1^{e}{\frac{1}{x}dx+\int_1^{e}{\frac{\log x}{x}dx}} $
therefore $ [{\log _{e}}x]_1^{e}+[ \frac{{{(\log x)}^{2}}}{2} ]_1^{e}=\frac{3}{2} $ .
BETA
