When doing this problem, let U equal sqrt 3 * 2 R -1 ^2 + 6. If we let that be our U, then our DU, we know that the derivative of a square root is 1 / 2 times that whole square root. And then by the chain rule, we have to take the derivative of the inside. So the derivative of the inside is going to be. We're going to bring down the two by the power rule, so two times the three. So we'll do times the derivative of the inside three times that two r -, 1 ^2 + 6. So bringing down the two 2 * 3 would give us 6 * 2 R -1 all to the one less power. So to the first times the derivative of the inside by the chain rule again times two, and then the derivative of that +6, the derivative of constants just zero. So we have all of that over that two square roots three times the quantity two r -, 1 ^2 + 6 and that's all times Dr. and that equals DU. So when we look at this and we simplify, the plus zero goes away. The two on top and the two on bottom are going to cancel. So we're going to get DU equaling 6 times the quantity 2 R -1 all over the square root, three times the quantity two r -, 1 ^2 + 6 Dr. So when we look at the original, we see that sqrt 3 times the quantity two r ^2 + 6 in the denominator and the Dr. and we need a six two R -1. We don't have the six, but we do have this two r -, 1. So if we look at this, we have all of this portion already in the original equation. The six isn't there. So we're going to take those six to the other side. I'm going to say 1/6 DU is 2 R -1 over the square root, three times the quantity two r -, 1 ^2 + 6 Dr. So we're going to take all of that out. We're going to replace it with 16 du, and then we have cosine of the square root of that, which is just U. So if we have the integral of 1/6 cosine U du, we know that that's going to be 1/6 sine U + C and then we're going to replace the sine with that square root three times the quantity two r -, 1 ^2 + 6 plus C.