If {I_m} = int_1^x {(log x)^m} dx satisfies t | vedclass

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(B) Given ${I_m} = \int_1^x {(\log x)^m} dx$. Using integration by parts,let $u = (\log x)^m$ and $dv = dx$. Then $du = m(\log x)^{m-1} \cdot \frac{1}

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$l = m$

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(B) Given ${I_m} = \int_1^x {(\log x)^m} dx$. Using integration by parts,let $u = (\log x)^m$ and $dv = dx$. Then $du = m(\log x)^{m-1} \cdot \frac{1}{x} dx$ and $v = x$. ${I_m} = [x(\log x)^m]_1^x - \int_1^x x \cdot m(\log x)^{m-1} \cdot \frac{1}{x} dx$. ${I_m} = x(\log x)^m - m \int_1^x (\log x)^{m-1} dx$. ${I_m} = x(\log x)^m - m{I_{m-1}}$. Comparing this with the given relation ${I_m} = k - l{I_{m-1}}$,we get $k = x(\log x)^m$ and $l = m$. Thus,the correct option is $l = m$.

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

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