Question:

Let t = \sqrt{x}, then x = t^2 and dx = 2t dt. The integral becomes \int(\sqrt{t^2+7}/t) * 2t dt = 2 \int \sqrt{t^2 + (\sqrt{7}}^2) dt. Using the formula \int\sqrt{t^2+a^2} dt = (t/2)\sqrt{t^2+a^2} + (a^2/2)ln|t+\sqrt{t^2+a^2}|, we get I(x) = t\sqrt{t^2+7} + 7ln|t+\sqrt{t^2+7}| + C = \sqrt{x}\sqrt{x+7} + 7ln|\sqrt{x}+\sqrt{x+7}| + C. Given I(9) = 3\sqrt{16} + 7ln(3+4) + C = 12 + 7ln7 + C = 12 + 7ln7, so C = 0. Then I(1) = 1\sqrt{8} + 7ln(1+\sqrt{8}) = 2\sqrt{2} + 7ln(1+2\sqrt{2}). Comparing with \alpha + 7ln(1+2\sqrt{2}), we get \alpha = 2\sqrt{2.} Thus \alpha^4 = (2\sqrt{2})^4 = 8^2 = 64.