int x^r lnx dx∫xrlnxdx is a 'standard' question.
For r!= -1r≠−1 integrate by parts.
We don't (most of us) know the integral of lnxlnx, but we do know its derivative, so
Let u = lnxu=lnx and dv = x^r dxdv=xrdx (in this question dv = x^(2/3) dxdv=x23dx
With these choices, we get:
du = 1/x dxdu=1xdx and v = int x^r dx = x^(r+1)/(r+1)v=∫xrdx=xr+1r+1 (Here: 3/5x^(5/3)35x53)
int u dv = uv - int vdu = 3/5 x^(5/3) lnx - 3/5 int x^(5/3) * 1/x dx∫udv=uv−∫vdu=35x53lnx−35∫x53⋅1xdx
= 3/5 x^(5/3) lnx - 3/5 int x^(2/3) dx=35x53lnx−35∫x23dx (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=35x53lnx−35⋅35x53+C Or for your definite integral:
= [3/5 x^(5/3) lnx - 9/25 x^(5/3)]_1^4=[35x53lnx−925x53]41
Now do the arithmetic. (remember that ln1 = 0ln1=0)
