Given Im=int1^e(logx)^mdx ,then prove t...

Given `I_m=int_1^e(logx)^mdx `,then prove that`(I_m)/(1-m)+m I_(m-2)=e`

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The correct Answer is:

`I_(m)=int_(1)^(e)(logx)^(m)dx`
`=(x(logx)^(m))_(1)^(e)-int_(1)^(e)x(m(logx)^(m-1))/xdx` (Integrating by parts)
`=e-m int_(1)^(e)(logx)^(m-1)dx=e-mI_(m-1)`…………..1
Replacing `m` by `m-1`, we get
`I_(m-1)=e-(m-1)I_(m-2)`…………..2
From 1 and 2 we have `I_(m)=e-m[e-(m-1)I_(m-2)]`
or `I_(m)-m(m-1)I_(m-2)=e(1-m)`
or `(I_(m)).(1-m)+mI_(m-2)=e`

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