These are completely equivalent statements. But this makes it clear that yes, u substitution will work over here. If we set our u equal to natural log of x, then our du is one over x dx, one over x dx. Our du is one over x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of one over u, natural log of x is u, we set that equal to natural log of x, times du, times du. Now this becomes pretty straightforward.