Biết

A

- Hướng dẫn giải

Đáp án A.

Cách 1:

$\begin{array}{l}\underset{1}{\overset{2}{\int }}\frac{2dx}{x^3+3x^2+2x}\\ =2.\underset{1}{\overset{2}{\int }}\frac{1}{x\left(x+1\right)\left(x+2\right)}dx\\ =\frac{2}{3}.\underset{1}{\overset{2}{\int }}\left(\frac{1}{x\left(x+2\right)}+\frac{1}{x\left(x+1\right)}-\frac{2}{\left(x+1\right)\left(x+2\right)}\right)dx\end{array}$

$=\frac{2}{3}\underset{1}{\overset{2}{\int }}\left(\frac{1}{2x}-\frac{1}{2x+4}+\frac{1}{x}-\frac{1}{x+1}-\frac{2}{x+1}+\frac{2}{x+2}\right)dx$

$$\begin{gathered} \underset{}{∲}=\frac{2}{3}\underset{1}{\overset{2}{\int }}\left(\frac{3}{2}.\frac{1}{x}-\frac{3}{x+1}+\frac{3}{2}.\frac{1}{x+2}\right)dx \\ =\underset{1}{\overset{2}{\int }}\frac{{\displaystyle 1}}{{\displaystyle x}}dx-2.\underset{1}{\overset{2}{\int }}\frac{{\displaystyle 1}}{{\displaystyle x+1}}dx+\underset{1}{\overset{2}{\int }}\frac{{\displaystyle 1}}{{\displaystyle x+2}}dx \end{gathered}$$

$\begin{array}{l}={\left.\ln \left|x\right|\right|}_1^2-{\left.2.\ln \left|x+1\right|\right|}_1^2+{\left.\ln \left|x+2\right|\right|}_1^2\\ =\ln 2-\ln 1-2.\left(\ln 3-\ln 2\right)+\ln 4-\ln 3\\ =\ln \frac{32}{27}\end{array}$

$\begin{array}{l}\Rightarrow a=32;\\ b=27\\ \Rightarrow a+b=59\end{array}$

Cách 2: Sử dụng máy tính.

Lúc này $\underset{1}{\overset{2}{\int }}\frac{2dx}{x^3+3x^2+2x}=Ans$  .

Suy ra $\frac{a}{b}=e^{Ans}$

$\Rightarrow \frac{a}{b}=\frac{32}{27}\Rightarrow a+b=59$