Giải thích các bước giải:

$\begin{array}{l}
10)\\
\int\limits_0^1 {\dfrac{{x + 1}}{{\sqrt {{x^2} + 2x + 2} }}} dx\\
 = \int\limits_0^1 {\dfrac{{2x + 2}}{{2\sqrt {{x^2} + 2x + 2} }}dx} \\
 = \int\limits_0^1 {\dfrac{{d\left( {{x^2} + 2x + 2} \right)}}{{2\sqrt {{x^2} + 2x + 2} }}} \\
 = \sqrt {{x^2} + 2x + 2} |_0^1\\
 = \sqrt 5  - \sqrt 2 \\
 \Rightarrow \left\{ \begin{array}{l}
a = 5\\
b = 2
\end{array} \right.\\
 \Rightarrow a - b = 5 - 2 = 3\\
 \Rightarrow D
\end{array}$

$\begin{array}{l}
11)\int\limits_1^a {\dfrac{{x + 1}}{x}dx} \\
 = \int\limits_1^a {\left( {1 + \dfrac{1}{x}} \right)dx} \\
 = \left( {x + \ln x} \right)|_1^a\\
 = a + \ln a - 1\\
 \Rightarrow a + \ln a - 1 = e\\
 \Leftrightarrow a + \ln a = e + \ln e\left( 1 \right)
\end{array}$

Xét hàm số $f\left( x \right) = x + \ln x,x \in \left( {0; + \infty } \right)$ 

Do $f'\left( x \right) = 1 + \dfrac{1}{x} > 0,\forall x \in \left( {0; + \infty } \right)$

$ \Rightarrow f\left( x \right) = 0$ có duy nhất một nghiệm

$ \Rightarrow \left( 1 \right)$ có nghiệm duy nhất $a = e$

$ \Rightarrow B$

$\begin{array}{l}
12)I = \int\limits_0^\pi  {\dfrac{{\sin x}}{{\sqrt {1 - 2a\cos x + {a^2}} }}dx} \\
 = \int\limits_0^\pi  {\dfrac{1}{a}.\dfrac{{2a\sin x}}{{2\sqrt {1 - 2a\cos x + {a^2}} }}dx} \\
 = \dfrac{1}{a}\int\limits_0^\pi  {\dfrac{{d\left( {1 - 2a\cos x + {a^2}} \right)}}{{2\sqrt {1 - 2a\cos x + {a^2}} }}} \\
 = \dfrac{1}{a}\sqrt {1 - 2a\cos x + {a^2}} |_0^\pi \\
 = \dfrac{1}{a}\left( {\sqrt {{a^2} + 2a + 1}  - \sqrt {{a^2} - 2a + 1} } \right)\\
 = \dfrac{1}{a}\left( {a + 1 - \left( {a - 1} \right)} \right)\\
 = \dfrac{2}{a}\\
 \Rightarrow A
\end{array}$

$ \Rightarrow \left( 1 \right)$