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Given $I_{m} = \int_{1}^{e} (\log x)^{m} d x$ , If $\frac{I_{m}}{K} + \frac{I_{m - 2}}{L} = e$ then values of K and L are

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Correct option is A

Im=∫1e (log⁡x)mdx=x(log⁡x)m1e−m∫1e (log⁡x)m−1dx=e−mx(log⁡x)m−11e−(m−1)∫1e (log⁡x)m−2dx=e−me+m(m−1)Im−2=(1−m)e+m(m−1)Im−2So Im1−m+mIm−2=e. Thus K=1−m and L=1m.

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Given Im=∫1e (log⁡x)mdx, If ImK+Im−2L= e then values of K and L are

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Given $I_{m} = \int_{1}^{e} (\log x)^{m} d x$ , If $\frac{I_{m}}{K} + \frac{I_{m - 2}}{L} = e$ then values of K and L are