#int x^r lnx dx# is a 'standard' question.
For #r!= -1# integrate by parts.
We don't (most of us) know the integral of #lnx#, but we do know its derivative, so
Let #u = lnx# and #dv = x^r dx# (in this question #dv = x^(2/3) dx#
With these choices, we get:
#du = 1/x dx# and #v = int x^r dx = x^(r+1)/(r+1)# (Here: #3/5x^(5/3)#)
#int u dv = uv - int vdu = 3/5 x^(5/3) lnx - 3/5 int x^(5/3) * 1/x dx#
#= 3/5 x^(5/3) lnx - 3/5 int x^(2/3) dx# (You'll always get the same integral fo this kind of problem.)
#= 3/5 x^(5/3) lnx - 3/5 *3/5 x^(5/3)+C# Or for your definite integral:
#= [3/5 x^(5/3) lnx - 9/25 x^(5/3)]_1^4#
Now do the arithmetic. (remember that #ln1 = 0#)
