Integral Calculus Question 189
Question: $ \int{\frac{{f(x).\phi ‘(x)-f’(x).\phi (x)}}{f(x).\phi (x)}}\log \frac{f(x)}{\phi (x)}dx $ is equal to:
Options:
A) $ \log \frac{\phi (x)}{f(x)}+k $
B) $ \frac{1}{2}{{{ \log \frac{\phi (x)}{f(x)} }}^{2}}+k $
C) $ \frac{\phi (x)}{f(x)}\log \frac{\phi (x)}{f(x)}+k $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ I=\int{\frac{f(x)\phi ‘(x)-f’(x)\phi (x)}{f(x)\phi (x)}\log ( \frac{\phi (x)}{f(x)} )dx} $ Putting $ \log \frac{\phi (x)}{f(x)}=t $ $ \frac{f(x)}{\phi (x)}.\frac{f(x)\phi ‘(x)-f’(x)\phi (x)}{{{(f(x))}^{2}}}.dx=dt, $ we get $ I=\int{tdt=\frac{1}{2}t^{2}+k=\frac{1}{2}{{( \log \frac{\phi (x)}{f(x)} )}^{2}}+k,k\in R} $
BETA
