Integral Calculus Question 41
Question: $ \int_{{}}^{{}}{\frac{dx}{x+x\log x}=} $
[MP PET 1993; Roorkee 1977]
Options:
A) $ \log (1+\log x) $
B) $ \log \log (1+\log x) $
C) $ \log x+\log (\log x) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{dx}{x+x\log x}}=\int_{{}}^{{}}{\frac{dx}{x(1+\log x)}} $ Now putting $ 1+\log x=t\Rightarrow \frac{1}{x},dx=dt, $ it reduces to $ \int_{{}}^{{}}{\frac{dt}{t}}=\log (t)=\log (1+\log x) $ .
BETA
