A) \[{{\log }_{e}}(1-{{\log }_{e}}x)+c\]
B) \[{{\log }_{e}}({{\log }_{e}}ex-1)+c\]
C) \[{{\log }_{e}}({{\log }_{e}}x-1)+c\]
D) \[{{\log }_{e}}({{\log }_{e}}x+x)+c\]
E) \[{{\log }_{e}}(1+{{\log }_{e}}x)+c\]
Correct Answer: E
Solution :
Let \[I=\int{\frac{1}{x}}({{\log }_{ex}}e)dx\] \[=\int{\frac{1}{x(1+{{\log }_{e}}x)}}dx\] Put \[{{\log }_{e}}x=t\Rightarrow \frac{1}{x}dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{(1+t)}}\] \[={{\log }_{e}}(1+t)+c\] \[={{\log }_{e}}(1+{{\log }_{e}}x)+c\]
You need to login to perform this action.
You will be redirected in
3 sec
