When doing this problem, we need to do a change of base because we don't know how to do log base three of X. So we're going to have DX6X instead of log base three of X. That's going to be natural log of N divided by natural log of three. We could take that natural log of three up to the top, remembering it's just a constant. So the natural log of 3 / 6 is really really just a constant. We still have the DX, and now we have an XL and X. If we don't remember, we can think about letting a equal log base three of X, changing it into exponential form SO3 to the A equal X, taking the natural log of each side, bringing down the exponent of the A in front, and then dividing each side. So we can see that A equals LNX divided by LN3. And originally we said A equal log base three of X, so by transitive property those two have to be equal. So once we have this, we still haven't evaluated the integral we're going to do AU sub. If we thought about letting U equal LNX, then DU would equal 1 over XDX. So now we can see we have this L and 3 / 6, which is just a constant. I actually will take it out in front of the integral because constants can come in and out. The one over XDX. This DX over X is really just going to be our DU and then we still have that L&X in the bottom and that's going to be U. So now when we integrate this, we get L and 3 / 6 the natural log of the absolute value of U + C and instead of U we need to substitute back. So we get natural log of 3 / 6 times the natural log of the absolute value of the natural log of X + C.