If In=∫(logx)ndx then I6+6I5=

In=∫(logx)ndx then,In=xlog xn-n In-1 ∵by reduction formula ⇒In+n In-1=xlog xn put n=6 ⇒I6+6 I5=xlog x6

If In=∫(logx)ndx then I6+6I5=

In=∫(logx)ndx then,In=xlog xn-n In-1 ∵by reduction formula ⇒In+n In-1=xlog xn put n=6 ⇒I6+6 I5=xlog x6

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If $I_{n} = \int {(\log x)}^{n} d x t h e n I_{6} + 6 I_{5} =$

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$x {(\log x)}^{5} + c$

$- x {(\log x)}^{5} + c$

$x {(\log x)}^{6} + c$

$- x {(\log x)}^{6} + c$

answer is C.

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Detailed Solution

$I_{n} = \int {(\log x)}^{n} d x then,$

$I_{n} = x {(\log x)}^{n} - n I_{n - 1} (∵ b y r e d u c t i o n f o r m u l a) \Rightarrow I_{n} + n I_{n - 1} = x {(\log x)}^{n} p u t n = 6 \Rightarrow I_{6} + 6 I_{5} = x {(\log x)}^{6}$

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If $I_{n} = \int {(\log x)}^{n} d x t h e n I_{6} + 6 I_{5} =$