$$\begin{array}{|c|c|c|}\hline \text{Câu hỏi}&\textbf{10}&\textbf{11}\\\hline \text{Đáp án}&\mathcal{B}&\mathcal{C}\\\hline\end{array}$$

$\textbf{10. }$

$$\begin{aligned}\displaystyle \int\dfrac{1}{x^2+x}\text{d}x &\quad=\displaystyle \int\dfrac{1}{x(x+1)}\text{d}x  \\ &\quad = \displaystyle \int\dfrac{1}{x}\text{d}x - \displaystyle \int\dfrac{1}{x+1}\text{d}x  \\ &\quad=\ln|x|-\ln|x+1|+\mathcal{C}  \\ &\quad=\ln\left|\dfrac{x}{x+1}\right|+\mathcal{C}  \\ &\quad \Longrightarrow \mathcal{B.} \end{aligned}$$

$\textbf{11. }$

$$\begin{aligned}\displaystyle \int\dfrac{1}{x^2-2x-3}\text{d}x &\quad=\displaystyle \int\dfrac{1}{(4x-12)(4x+4)}\text{d}x  \\ &\quad = \dfrac{1}{4}\displaystyle \int\dfrac{1}{x-3}\text{d}x - \dfrac{1}{4} \displaystyle \int\dfrac{1}{x+1}\text{d}x  \\ &\quad=\dfrac{1}{4}\left(\ln|x-3|-\ln|x+1| \right)+\mathcal{C}  \\ &\quad=\dfrac{1}{4}\ln\left|\dfrac{x-3}{x+1}\right|+\mathcal{C}  \\ &\quad \Longrightarrow \mathcal{C.} \end{aligned}$$

$@thomasnguyen364$