\[I=\int{\left\{ {{\log }_{e}}{{\log }_{e}}x+\frac{1}{{{({{\log }_{e}}x)}^{2}}} \right\}dx}\] is equal to:

\[x{{\log }_{e}}{{\log }_{e}}x+\frac{x}{{{\log }_{e}}x}+c\]

\[x{{\log }_{e}}{{\log }_{e}}x+c\]

\[x{{\log }_{e}}{{\log }_{e}}x-\frac{x}{{{\log }_{e}}x}+c\]

\[I=\int{\left\{ {{\log }_{e}}{{\log }_{e}}x+\frac{1}{{{({{\log }_{e}}

1.	\[I=\int{\left\{ {{\log }_{e}}{{\log }_{e}}x+\frac{1}{{{({{\log }_{e}}x)}^{2}}} \right\}dx}\] is equal to:
A.	\[x{{\log }_{e}}{{\log }_{e}}x+c\]
B.	\[x{{\log }_{e}}{{\log }_{e}}x-\frac{x}{{{\log }_{e}}x}+c\]
C.	\[x{{\log }_{e}}{{\log }_{e}}x+\frac{x}{{{\log }_{e}}x}+c\]
D.	None of these.
Answer» C. \[x{{\log }_{e}}{{\log }_{e}}x+\frac{x}{{{\log }_{e}}x}+c\]

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\[I=\int{\left\{ {{\log }_{e}}{{\log }_{e}}x+\frac{1}{{{({{\log }_{e}}x)}^{2}}} \right\}dx}\] is equal to:

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