Giả sử

A

- Hướng dẫn giải

Đáp án A

Ta có

$F\left(x\right)=-\int \ln \left(x+3\right)d\left(\frac{1}{x}\right)$$=-\frac{\ln \left(x+3\right)}{x}+\int \frac{1}{x}d\left[\ln \left(x+3\right)\right]$

$=-\frac{\ln \left(x+3\right)}{x}+\int \frac{1}{x}.\frac{1}{x+3}d\text{x}$$=-\frac{\ln \left(x+3\right)}{x}+\frac{1}{3}\int \left(\frac{1}{x}-\frac{1}{x+3}\right)d\text{x}$

$=-\frac{\ln \left(x+3\right)}{x}+\frac{1}{3}\ln \left|\frac{x}{x+3}\right|+C.$

Mà 

$\begin{array}{l}F\left(-2\right)+F\left(1\right)=0\\ \Rightarrow \left(\frac{1}{3}\ln 2+C_1\right)+\left(-\ln 4+\frac{1}{3}\ln \frac{1}{4}+C_2\right)=0\\ \Rightarrow -\frac{7}{3}\ln 2+C_1+C_2=0\end{array}$

$\begin{array}{l}\Rightarrow F\left(-1\right)+F\left(2\right)\\ =\left(\ln 2+\frac{1}{3}\ln \frac{1}{2}+C_1\right)+\left(-\frac{1}{2}\ln 5+\frac{1}{3}\ln \frac{2}{5}+C_2\right)\\ =\frac{10}{3}\ln 2-\frac{5}{6}\ln 5.\end{array}$