Integral Calculus Question 350
Question: The value of $ \int{\frac{\log x}{{{(x+1)}^{2}}}dx} $ is
[UPSEAT 1999]
Options:
A) $ \frac{-\log x}{x+1}+\log x-\log ,(x+1) $
B) $ \frac{\log x}{( x+1 )}+\log x-\log ,(x+1) $
C) $ \frac{\log x}{x+1}-\log x-\log ,(x+1) $
D) $ \frac{-\log x}{x+1}-\log x-\log ,(x+1) $
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Answer:
Correct Answer: A
Solution:
$ \int{\frac{\log x}{{{(x+1)}^{2}}}dx=\int{\log x,{{(x+1)}^{-2}}}}dx $ $ =\log x.{ -{{(x+1)}^{-1}} } $ $ -\int{\frac{1}{x}.{-{{(x+1)}^{-1}}}dx} $ $ =\frac{-\log x}{(x+1)}+\int{\frac{1}{x(x+1)}dx} $ $ =\frac{-\log x}{(x+1)}+\int{[ \frac{1}{x}-\frac{1}{x+1} ]dx} $ $ =\frac{-\log x}{x+1}+\log x-\log (x+1) $ .
BETA
