Ghi chú: Công thức nguyên hàm từng phần.
$$\displaystyle \int \limits_{\alpha}^{\beta} u \text{d}v = uv \Biggr|_{\alpha}^{\beta} - \displaystyle \int \limits_{\alpha}^{\beta} v \text{d}u$$

$$\displaystyle\left \{ {{u=x} \atop {\text{d}v=\dfrac{\text{d}x}{(x+2)^2}}} \right. \Rightarrow \left \{ {{\text{d}u=\text{d}x} \atop {v=-\dfrac{1}{x+2}}} \right.$$
$$\begin{aligned} \mathcal{I}&= \displaystyle \int \limits_{0}^{1} \dfrac{x}{(x+2)^2} \text{d}x 
\\ &=\left( -\dfrac{x}{x+2}\right)\Biggr|_{0}^{1} + \displaystyle \int \limits_{0}^{1} \dfrac{\text{d}x}{x+2}
\\ &= -\dfrac{1}{3}+ \ln{\left|x+2\right|} \Biggr|_{0}^{1} 
\\ &= -\dfrac{1}{3}+ \ln{3} - \ln{2} 
\\ & \Rightarrow \displaystyle \begin{cases}
\textsf{a} =-\dfrac{1}{3}\\
\textsf{b} = -1\\
\textsf{c} = 1
\end{cases}
\end{aligned}$$

$\Leftrightarrow 3 \textsf{a} + \textsf{b} + \textsf{c} = -\dfrac{3}{3}-1+1=-1$

$@thomasnguyen364$