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

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.