Đáp án:
$\begin{array}{l}
A = \int\limits_1^2 {\frac{1}{{\sqrt x }}dx} \\
= \int\limits_1^2 {2.\frac{1}{{2\sqrt x }}dx} \\
= \left( {2\sqrt x } \right)_1^2 = 2\sqrt 2 - 2\\
B = \int\limits_1^2 {\frac{1}{{\sqrt {x + 1} }}dx} \\
= \int\limits_1^2 {2.\frac{1}{{2\sqrt {x + 1} }}d\left( {x + 1} \right)} \\
= \left( {2\sqrt {x + 1} } \right)_1^2\\
= 2\sqrt 3 - 2\sqrt 2 \\
\Rightarrow I = A - B\\
= \left( {2\sqrt 2 - 1} \right) - \left( {2\sqrt 3 - 2\sqrt 2 } \right)\\
= 4\sqrt 2 - 2\sqrt 3 - 1
\end{array}$
