A) \[{{e}^{10}}\]
B) \[\frac{{{e}^{10}}}{10}\]
C) \[e\]
D) \[e-1\]
Correct Answer: C
Solution :
We have, |
\[{{I}_{n}}=\int\limits_{1}^{e}{{{(\ln x)}^{n}}dx}\]\[\Rightarrow \]\[{{I}_{n}}=[{{(\ln x)}^{n}}x]_{1}^{e}-\int\limits_{1}^{e}{\frac{n\,{{(\ln x)}^{n-1}}}{x}}\cdot x\,dx\]\[\Rightarrow \]\[{{I}_{n}}=e-n{{I}_{n-1}}\]\[\Rightarrow \]\[{{I}_{n}}+n{{I}_{n-1}}=e\] |
\[\therefore \]\[{{I}_{10}}+10{{I}_{9}}=e\] |
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