Integration using part formula \int\limits {x^nlogx} \, dx

Question

Integration using part formula

`\int\limits {x^nlogx} \, dx`

Answer 1

Integration of I=
`\int {x^(n)logx \, dx`=
`[(logx)/(n+1) x^((n+1))]-[(1)/((n+1)^(2))x^((n+1))]`

Explanation:

Given integral is I=
`\int {x^(n)logx \, dx`

Take logx=t

`x=e^(t)`

`x^(n)=e^(nt)`

`(1)/(x) dx=dt`

`dx=xdt`

`dx=e^(t)dt`

I=
`\int (e^(nt))(t)(e^(t))\, dt`

I=
`\int (e^((n+1)t))(t)\, dt`

Using integration by part,

`I= (t)\int [e^((n+1)t)]\, dt-\int[(d)/(dt){t}*\int (e^((n+1)t))]\\\\I= (t) [(1)/(n+1)e^((n+1)t)]-\int[1*(1)/(n+1)e^((n+1)t)]\,dt\\\\I=[(t)/(n+1)e^((n+1)t)]-[(1)/((n+1)^(2))e^((n+1)t)]`

Writing in terms of x

I=
`[(t)/(n+1)e^((n+1)t)]-[(1)/((n+1)^(2))e^((n+1)t)]`

I=
`[(logx)/(n+1)e^((n+1)logx)]-[(1)/((n+1)^(2))e^((n+1)logx)]`

I=
`[(logx)/(n+1)e^{logx^((n+1))}]-[(1)/((n+1)^(2))e^{logx^((n+1))}]`

I=
`[(logx)/(n+1) x^((n+1))]-[(1)/((n+1)^(2))x^((n+1))]`

Thus,

Integration of I=
`\int {x^(n)logx \, dx`=
`[(logx)/(n+1) x^((n+1))]-[(1)/((n+1)^(2))x^((n+1))]`