If I_m = _1^x ( x) ^m dx satisfies the relation I_m = k - l I_ m … - Vidyadip

MCQ

If ${I_m} = \int_1^x {{{(\log x)}^m}dx} $ satisfies the relation ${I_m} = k - l{I_{m - 1}},$ then

A
$k = e$
✓
$l = m$
C
$k = \frac{1}{e}$
D
None of these

✓

Answer

Correct option: B.

$l = m$

b
(b) ${I_m} = \int_1^x {{{(\log x)}^m}dx} $

$= ({(\log x)^m}.x)_1^x - \int_1^x {m{{(\log x)}^{m - 1}}.\frac{1}{x}x\,\,dx} $

$ = {(\log x)^m}.x - m{I_{m - 1}}$

$\therefore \,\,$${I_m} = k - l\,\,{I_{m - 1}} $

$\Rightarrow k - l\,{I_{m - 1}} = x{(\log x)^m} - m{I_{m - 1}}$

==> $k = x{(\log x)^m},l = m$.

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