If In-log xn d x- then In-n In-1-

The correct option is A x(log x)^n I_n =∫ (log x)^n dx....(i) ∴ I_n 1=∫ (log x)^n 1dx ....(ii) Now, I_n =∫ (log x)^n.dx=(log x)^n x n ∫ (log x)^n 11/xx dx =x ...

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Byju's Answer

Standard XII

Mathematics

Progression

If In=∫log xn...

Question

If $I_{n} = \int (l o g x)^{n} d x$ , then $I_{n} + n I_{n - 1} =$

x (l o g x)^{n}

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(x l o g x)^{n}

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(l o g x)^{n - 1}

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n (l o g x)^{n}

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Solution

The correct option is A $x (l o g x)^{n}$

$I_{n} = \int (l o g x)^{n} d x . . . . (i) ∴ I_{n - 1} = \int (l o g x)^{n - 1} d x . . . . (i i)$
Now, $I_{n} = \int (l o g x)^{n} . d x = (l o g x)^{n} x - n \int (l o g x)^{n - 1} \frac{1}{x} x d x$
$= x (l o g x)^{n} - n \int (l o g x)^{n - 1} d x I_{n} = x (l o g x)^{n} - n I_{n - 1}; ∴ I_{n} + n I_{n - 1} = x (l o g x)^{n}$

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If In=∫(logx)ndx, then In+nIn−1=

The correct option is A x(logx)n In=∫(logx)ndx....(i)∴In−1=∫(logx)n−1dx....(ii) Now, In=∫(logx)n.dx=(logx)nx−n∫(logx)n−11xx dx =x(logx)n−n∫(logx)n−1dxIn=x(logx)n−nIn−1;∴In+nIn−1=x(logx)n

If $I_{n} = \int (l o g x)^{n} d x$ , then $I_{n} + n I_{n - 1} =$