To evaluate this integral, we're going to split hyperbolic cotangent into hyperbolic cosine divided by hyperbolic sine, and we're going to do AU sub. So if I let U equal the hyperbolic sine of 6X, then our DU is 6 hyperbolic cosine 6 XDX or 16 DU equals hyperbolic cosine 6 XDX. So when we substitute here, that hyperbolic cosine 6 XDX just gets replaced with 160 U, and that leaves me one over U. If I go ahead and change my bounds, I get U to the LN5, so that's going to equal the hyperbolic sine of six times LN5. Using my definition, I get E to the LN5 to the 6th minus E to the LN5 to the -6 all over 2. The E to the LNS cancels, so 5 to the 6th -1 / 5 to the 6th all over two doing the bottom bound U of LN3. Same exact skills, hyperbolic sine of six times Ln threes. Take the six up in front. Put it into the official definition. So when I evaluate that integral, we get 1/6 the natural log of the absolute value of U sticking in the upper bound, then minus the lower bound. That gives us approximately .5108 direction, say round to the nearest hundredth, so .51.