If I_n = ( x) ^n , ,dx , then I_n + n I_ n - 1 = - Vidyadip

MCQ

If ${I_n} = \int {{{(\log x)}^n}\,\,dx} ,$ then ${I_n} + n{I_{n - 1}} = $

✓
$x{(\log x)^n}$
B
${(x\log x)^n}$
C
${(\log x)^{n - 1}}$
D
$n{(\log x)^n}$

✓

Answer

Correct option: A.

$x{(\log x)^n}$

a
(a)${I_n} = \int {{{(\log x)}^n}dx} $ .....$(i)$
$\therefore {I_{n - 1}} = \int {{{(\log x)}^{n - 1}}dx} $.....$(ii)$
Now, ${I_n} = \int {{{(\log x)}^n}.\,dx} = {(\log x)^n}x - n\int {{{(\log x)}^{n - 1}}\frac{1}{x}x\,dx} $
$ = x{(\log x)^n} - n\int {{{(\log x)}^{n - 1}}dx} $
${I_n} = x{(\log x)^n} - n{I_{n - 1}}$; $\therefore {I_n} + n\,{I_{n - 1}} = x{(\log x)^n}$.

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