\(F\left(x\right)=\int\dfrac{ln\left(x+3\right)}{x^2}dx\)
Đặt \(\left\{{}\begin{matrix}u=ln\left(x+3\right)\\dv=\dfrac{dx}{x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x+3}\\v=-\dfrac{1}{x}\end{matrix}\right.\)
\(\Rightarrow F\left(x\right)=-\dfrac{ln\left(x+3\right)}{x}+\int\dfrac{dx}{x\left(x+3\right)}=-\dfrac{ln\left(x+3\right)}{x}+\dfrac{1}{3}\int\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right)dx\)
\(=-\dfrac{ln\left(x+3\right)}{x}+\dfrac{1}{3}ln\left|\dfrac{x}{x+3}\right|+C\)
\(F\left(-2\right)+F\left(1\right)=\dfrac{1}{3}ln2-ln4+\dfrac{1}{3}ln\left(\dfrac{1}{4}\right)+C\)
\(=\dfrac{1}{3}ln2-2ln2-\dfrac{2}{3}ln2+C=0\Rightarrow C=\dfrac{7}{3}ln2\)
\(\Rightarrow F\left(x\right)=-\dfrac{ln\left(x+3\right)}{x}+\dfrac{1}{3}ln\left|\dfrac{x}{x+3}\right|+\dfrac{7}{3}ln2\)
