A) \[\frac{n}{n+1}{{I}_{m,n}}_{-1}\]
B) \[\frac{-m}{n+1}{{I}_{m,n-1}}\]
C) \[\frac{-n}{m+1}{{I}_{m,n-1}}\]
D) None of the above
Correct Answer: C
Solution :
\[{{I}_{m,n}}=\int_{0}^{1}{{{x}^{m}}{{(\log x)}^{n}}dx}\] \[=\left[ {{(\log x)}^{n}}.\frac{{{x}^{m+1}}}{m+1} \right]_{0}^{1}\] \[-\int_{0}^{1}{n{{(\log x)}^{n-1}}.\frac{1}{x}.\frac{{{x}^{m+1}}}{(m+1)}dx}\] \[=0-\frac{n}{(m+1)}\int_{0}^{1}{{{x}^{m}}{{(\log x)}^{n-1}}dx}\] \[=\frac{-n}{m+1}.{{I}_{m,n-1}}\]
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