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
B
nm+1Im,n1
C
1m+1Im,n1
D
mn+1Im,n1

Solution

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

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