Tích phân $I=\displaystyle\int\limits_{1}^{2} \dfrac{1}{\sqrt{x+1}-\sqrt{x}}\mathrm{d}x$ bằng

Ta có: $\dfrac{1}{\sqrt{x+1} - \sqrt{x}} = \dfrac{\sqrt{x+1} + \sqrt{x}}{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})} = \dfrac{\sqrt{x+1} + \sqrt{x}}{x+1 - x} = \sqrt{x+1} + \sqrt{x}$
Do đó: $I = \displaystyle\int\limits_{1}^{2} (\sqrt{x+1} + \sqrt{x}) dx = \displaystyle\int\limits_{1}^{2} (x+1)^{\frac{1}{2}} dx + \displaystyle\int\limits_{1}^{2} x^{\frac{1}{2}} dx$
$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3^{\frac{3}{2}} - 2^{\frac{3}{2}}) + \dfrac{2}{3}(2^{\frac{3}{2}} - 1^{\frac{3}{2}}) = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) + \dfrac{2}{3}(2\sqrt{2} - 1) = 2\sqrt{3} - \dfrac{2}{3} + \dfrac{4\sqrt{2}}{3} - \dfrac{4\sqrt{2}}{3} = 2\sqrt{3} + \dfrac{6}{3} - \dfrac{2}{3} = 2\sqrt{3} + \dfrac{4}{3}$
$I = \dfrac{2}{3} ( (x+1)\sqrt{x+1} )\Big|_1^2 + \dfrac{2}{3} ( x\sqrt{x} )\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) + \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3} = 2\sqrt{3} - \dfrac{2}{3}$
Tuy nhiên không có đáp án nào trùng khớp, ta kiểm tra lại đề bài, ta thấy có lẽ dấu trừ trong mẫu số phải là dấu cộng
$I = \displaystyle\int\limits_{1}^{2} \dfrac{1}{\sqrt{x+1}+\sqrt{x}}\mathrm{d}x = \displaystyle\int\limits_{1}^{2} (\sqrt{x+1} - \sqrt{x}) dx = \displaystyle\int\limits_{1}^{2} (x+1)^{\frac{1}{2}} dx - \displaystyle\int\limits_{1}^{2} x^{\frac{1}{2}} dx$
$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 - \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3^{\frac{3}{2}} - 2^{\frac{3}{2}}) - \dfrac{2}{3}(2^{\frac{3}{2}} - 1^{\frac{3}{2}}) = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) - \dfrac{2}{3}(2\sqrt{2} - 1) = 2\sqrt{3} - \dfrac{2}{3} - \dfrac{4\sqrt{2}}{3} + \dfrac{4\sqrt{2}}{3} = 2\sqrt{3} + \dfrac{6}{3} - \dfrac{2}{3} = 2\sqrt{3} - \dfrac{4}{3}$
$I = \dfrac{2}{3} ( (x+1)\sqrt{x+1} )\Big|_1^2 - \dfrac{2}{3} ( x\sqrt{x} )\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) - \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3} = 2\sqrt{3} + \dfrac{2}{3}$
Vậy đáp án đúng phải là $2\sqrt{3} + \dfrac{2}{3}$
Nếu đề là $I=\displaystyle\int\limits_{1}^{2} (\sqrt{x+1}+\sqrt{x}) dx$ thì ta làm như sau:
$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) + \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3} = 2\sqrt{3} - \dfrac{2}{3}$
Nếu đề là $I=\displaystyle\int\limits_{1}^{2} (\sqrt{x+1}-\sqrt{x}) dx$ thì ta làm như sau:
$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 - \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) - \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3} = 2\sqrt{3} + \dfrac{2}{3}$
Để ý nếu đổi cận từ 1->2 thành 0->1 thì:
$I=\displaystyle\int\limits_{0}^{1} (\sqrt{x+1}-\sqrt{x}) dx = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_0^1 - \dfrac{2}{3}x^{\frac{3}{2}}\Big|_0^1 = \dfrac{2}{3}(2\sqrt{2} - 1) - \dfrac{2}{3}(1 - 0) = \dfrac{4\sqrt{2}}{3} - \dfrac{4}{3}$
Nếu đổi cận từ 1->2 thành 0->1 thì:
$I=\displaystyle\int\limits_{0}^{1} (\sqrt{x+1}+\sqrt{x}) dx = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_0^1 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_0^1 = \dfrac{2}{3}(2\sqrt{2} - 1) + \dfrac{2}{3}(1 - 0) = \dfrac{4\sqrt{2}}{3}$