Question:
If $I_{m}=\int\limits_{1}^{e}(\ln x)^{m} d x$, where $m \in N$, then $I_{10}+10 I_{9}$ is equal to -
If $I_{m}=\int\limits_{1}^{e}(\ln x)^{m} d x$, where $m \in N$, then $I_{10}+10 I_{9}$ is equal to -
Updated On: Apr 14, 2026
- $e^{10}$
- $\frac{e^{10}}{10}$
- $e$
- $e^{ -1}$
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The Correct Option is C
Solution and Explanation
$I_{10}=\int\limits_{1}^{e} 1 .(\ln x)^{10} d x=\left[(\ln x)^{10} x\right]_{1}^{e}$
$-\int\limits_{-1}^{e} 10(\ln x)^{9} \cdot \frac{1}{x} \cdot x d x$
$=e-0-10 \int\limits_{1}^{e}(\ln x)^{9} d x$
$=e-10 I_{9}+10 I_{9}=e$
$-\int\limits_{-1}^{e} 10(\ln x)^{9} \cdot \frac{1}{x} \cdot x d x$
$=e-0-10 \int\limits_{1}^{e}(\ln x)^{9} d x$
$=e-10 I_{9}+10 I_{9}=e$
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