When x=1, ln(x)=0 so 1/(xln3(x))→∞. When x=0, ln(x)→-∞. When x=1/e (for example), ln(1/e)=-1=ln3(1/e). So for 0<x<1, the expression is negative with asymptotes x=0 (y axis) and x=1.
Let y=ln(x), dy=dx/x. When x=0, y→-∞; when x→∞, y→0.
The indefinite integral becomes:
∫y-3dy=-1/(2y2)=-1/(2ln2(x))(+C). The definite integral can be expressed as:
[-1/(2ln2(x))]01+[-1/(2ln2(x))]1∞. The first term is a negative area (below the x axis). The second term is above the x axis.
Because the first term "starts" at -∞ and "ends" at -∞, there must be a maximum somewhere between x=0 and x=1.
Let z=1/(xln3(x)), then z=(d/dx)(x-1ln-3(x))=-1/(x2ln3(x))-3/(x2ln4(x))=0 at maximum.
-1/(x2ln3(x))=3/(x2ln4(x)), ln(x)=-3, x=e-3=0.0498 approx, z=-1/(27x)=-0.7439 approx.
[-1/(2ln2(x))]01=[-1/(2ln2(x))]0e⁻³+[-1/(2ln2(x))]e⁻³1=(-1/18+0)+(-∞+1/18)→-∞. This is an infinite area below the x axis.
The second term evaluates to -0+∞→∞. So the total area is undefined (unbounded, infinite).
This would seem to be the case if z is shown graphically. The area between z and the x-axis on both the positive and negative sides is unbounded. It can be argued that the two "infinities" (above and below the x-axis) effectively "cancel out", because areas below the x axis are evaluated as negative. It can also be argued that the two infinities are different in nature and cannot be treated as numbers, making the area indefinite. What is interesting is that there appears to be a finite area (1/18 below the x axis) between x=0 and x=e-3, implying a convergence.