Integral Calculus Question 97
Question: The value of $ \int\limits_1^{e}{\frac{1+x^{2}\ln ,x}{x+x^{2}\ln ,x}dx} $ is
Options:
A) $ e $
B) $ \ln ,(1+e) $
C) $ e+\ln (1+e) $
D) $ e-\ln (1+e) $
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Answer:
Correct Answer: D
Solution:
[d] $ I=\int\limits_1^{e}{( \frac{1}{x}+1 )}dx-\int\limits_1^{e}{\frac{1+\ln x}{1+x\ln x}dx} $ $ ={{[\ln x+x]}_1}^{e}-[\ln (1+x\ln x)]_1^{e} $ $ =e-\ln (1+e) $
BETA
