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Question

If $${ I }_{ m,n }=\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ m }{ \left( \log { x }  \right)  }^{ n }dx } $$, then $${I}_{m,n}$$it is equal to


A
nn+1Im,n1
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B
nm+1Im,n1
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C
1m+1Im,n1
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D
mn+1Im,n1
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Solution

The correct option is C $$- \dfrac { 1 }{ m+1 } { I }_{ m,n-1 }$$
$${ I }_{ m,n }=\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ m }{ \left( \log { x }  \right)  }^{ n }dx } $$
$$={ \left[ { \left( \log { x }  \right)  }^{ n }\cdot \dfrac { { x }^{ m+1 } }{ m+1 }  \right]  }_{ 0 }^{ 1 }-\displaystyle\int _{ 0 }^{ 1 }{ n{ \left( \log { x }  \right)  }^{ n-1 }\cdot \dfrac { 1 }{ x } \cdot \dfrac { { x }^{ m+1 } }{ \left( m+1 \right)  } dx } $$
$$=0-\dfrac { n }{ \left( m+1 \right)  } \displaystyle\int _{ 0 }^{ 1 }{ { x }^{ m }{ \left( \log { x }  \right)  }^{ n-1 }dx } $$ 
$$=\dfrac { -n }{ m+1 } \cdot { I }_{ m,n-1 }$$

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