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

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 #