Integral Calculus Question 460
Question: $ \int_{{}}^{{}}{\frac{1}{x}\log x\ dx} $ is equal to
[SCRA 1996]
Options:
A) $ \frac{1}{2}\log x+c $
B) $ \frac{1}{2}{{(\log x)}^{2}}+c $
C) $ \frac{1}{2}\log {{(x)}^{2}}+c $
D) $ \log x+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_{{}}^{{}}{\frac{1}{x}\log x,dx} $ . Put $ \log x=t\Rightarrow \frac{1}{x},dx=dt $
$ \therefore ,I\int_{{}}^{{}}{t,dt}=\frac{t^{2}}{2}+c=\frac{{{(\log x)}^{2}}}{2}+c $ .
BETA
