Question

Evaluate the integral? : `int_1^e x^5 (lnx)^2 dx`

Answer 1

`int_1^e \ x^5 \ (lnx)^2 \ dx = (13e^6 -1)/108`

Explanation:

We want to find:

`I = int_1^e \ x^5 \ (lnx)^2 \ dx`

We can use integration by Parts

Let `{ (u,=(lnx)^2, => (du)/dx,=(2lnx)/x), ((dv)/dx,=x^5, => v,=x^6/6 ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int_1^e \ (lnx)^2(x^5) \ dx = [(lnx)^2(x^6/6)]_1^e - int_1^e \ (x^6/6)((2lnx)/x) \ dx`

`:. I = 1/6(e^6ln^2e - ln^21) - 1/3 \ int_1^e \ x^5lnx \ dx`
`" "= e^6/6 - 1/3 \ int_1^e \ x^5lnx \ dx`

So now let us look at this next integral:

`J = int_1^e \ x^5lnx \ dx`

Again we can use integration by Parts

Let `{ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=x^5, => v,=x^6/6 ) :}`

and plugging into the IBP formula gives us

`int_1^e \ (lnx)(x^5) \ dx = [(lnx)(x^6/6)]_1^e - int_1^e \ (x^6/6)(1/x) \ dx`

`:. J = 1/6(e^6lne-ln1) - 1/6 \ int_1^e \ x^5 \ dx`
`" " = e^6/6 - 1/6 \ [x^6/6]_1^e`
`" " = e^6/6 - 1/36 \ (e^6-1)`

Combining these results we get:

`I = e^6/6 - 1/3 (e^6/6 - 1/36 \ (e^6-1))`
`\ \ = e^6/6 - 1/3 (e^6/6 - e^6/36+1/36)`
`\ \ = e^6/6 - 1/3 (5/36e^6 +1/36)`
`\ \ = e^6/6 -5/108 e^6 -1/108`
`\ \ = 13/108 e^6 -1/108`
`\ \ = (13e^6 -1)/108`