Câu hỏi:

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:

$I=\displaystyle\int\limits_{0}^{1} (\sqrt{x+1}+\sqrt{x}) dx = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_0^1 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_0^1 = \dfrac{2}{3}(2\sqrt{2} - 1) + \dfrac{2}{3}(1 - 0) = \dfrac{4\sqrt{2}}{3}$

Ta có: $I=\displaystyle\int\limits_{1}^{2}{\dfrac{1}{2}}\Big(\dfrac{1}{x}-\dfrac{1}{x+2} \Big)\mathrm{d}x = \dfrac{1}{2} \displaystyle\int\limits_{1}^{2} \Big(\dfrac{1}{x}-\dfrac{1}{x+2} \Big)\mathrm{d}x $
$\quad = \dfrac{1}{2} \Big[ \ln |x| - \ln |x+2| \Big]_{1}^{2} = \dfrac{1}{2} \Big[ \ln \Big|\dfrac{x}{x+2}\Big| \Big]_{1}^{2}$
$\quad = \dfrac{1}{2} \Big( \ln \dfrac{2}{4} - \ln \dfrac{1}{3} \Big) = \dfrac{1}{2} \Big( \ln \dfrac{1}{2} - \ln \dfrac{1}{3} \Big) $
$\quad = \dfrac{1}{2} (\ln 1 - \ln 2 - \ln 1 + \ln 3) = \dfrac{1}{2} (-\ln 2 + \ln 3) = -\dfrac{1}{2} \ln 2 + \dfrac{1}{2} \ln 3$
$\Rightarrow a = -\dfrac{1}{2}, b = \dfrac{1}{2}$
Vậy $T = a^2 + b^3 = \Big(-\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^3 = \dfrac{1}{4} + \dfrac{1}{8} = \dfrac{3}{8}$

Ta có: $\dfrac{x-2}{x+1} = 1 - \dfrac{3}{x+1}$.
Xét dấu: $x \in [1;2)$ thì $\dfrac{x-2}{x+1} < 0$; $x \in (2;5]$ thì $\dfrac{x-2}{x+1} > 0$.
Do đó:
$\displaystyle \int\limits_1^5 \left| \dfrac{x-2}{x+1} \right|\mathrm{d}x = -\int\limits_1^2 \dfrac{x-2}{x+1} dx + \int\limits_2^5 \dfrac{x-2}{x+1} dx$
$\displaystyle = -\int\limits_1^2 (1 - \dfrac{3}{x+1}) dx + \int\limits_2^5 (1 - \dfrac{3}{x+1}) dx$
$\displaystyle = -\left[ x - 3\ln|x+1| \right]_1^2 + \left[ x - 3\ln|x+1| \right]_2^5$
$\displaystyle = -\left[ (2 - 3\ln 3) - (1 - 3\ln 2) \right] + \left[ (5 - 3\ln 6) - (2 - 3\ln 3) \right]$
$\displaystyle = -1 + 3\ln 3 + 3\ln 2 + 3 - 3\ln 6$
$\displaystyle = 2 + 3\ln (\dfrac{6}{6}) = 2 + 3\ln 1 = 2$
Suy ra $a = 2, b = 0, c = 0$. Vậy $P = abc = 0$.

Ta có:
$\displaystyle\int\limits_{1}^{2}\big[ f(x)+4x^3 \big]\mathrm{d}x = \displaystyle\int\limits_{1}^{2} f(x) \mathrm{d}x + \displaystyle\int\limits_{1}^{2} 4x^3 \mathrm{d}x$

$= 3 + x^4 \Big|_1^2 = 3 + (2^4 - 1^4) = 3 + (16 - 1) = 3 + 15 = 18$.

Ta có:
$\displaystyle\int\limits_{-3}^{1}(2x-5)\mathrm{d}x = (x^2 - 5x) \Big|_{-3}^{1} = (1^2 - 5(1)) - ((-3)^2 - 5(-3)) = (1-5) - (9+15) = -4 - 24 = -28$.