When doing this type of problem, we can see that we have it in the form of a Pythagorean theorem. A ^2 + b ^2 = C ^2. So we have the a ^2 + b ^2 part. So if we make our right triangle, we're going to put our Theta here and we're going to have the opposite side be two U because two U * 2 U is 4U squared and the adjacent side be 9. The reason we do this is so that we can have tangent Theta equaling the two U / 9. If we take the derivative of each side, we'd get secant squared Theta D Theta equaling 2 ninths DU. Or we could think of that as nine halves, secant squared, Theta, D, Theta, equaling DU. So now when we come back to here, the four in the numerator can just come out in front because it's a constant. And our nine halves can go out in front for the same reason. It's a constant. So that DU is going to turn into the nine halves and we're going to have secant squared Theta, D Theta and the top and the bottom we have 81 and instead of 4U squared, come back here and think about nine tangent Theta equals 2 U. So if we square each side, we would get 81 tangent squared Theta. So the nine halves and the four are going to simplify to give us 18, the secant squared Theta, D Theta, the denominator. If I pull out an 81, I get one plus tangent squared Theta left. That's really just secant squared Theta. We can pull out the 81. So 1880 first secant squared Theta, D Theta. That one plus tangent squared Theta is secant squared Theta 18 / 81, both divisible by 9. So we're going to get 2 ninths. The secants are going to cancel. So the integral of D Theta is Theta plus C but Theta from back here we could have being tangent inverse of two U / 9 so that Theta turns into tangent inverse two U / 9 + C.