Lời giải:Ta có \(f\left( x \right)+g\left( x \right)=-x\left[ {f}'\left( x \right)+{g}'\left( x \right) \right]\)
\(\Rightarrow \int{\left[ f\left( x \right)+g\left( x \right) \right]\text{d}x}=-\int{x\left[ {f}'\left( x \right)+{g}'\left( x \right) \right]\text{d}x}\)
\(\Rightarrow \int{\left[ f\left( x \right)+g\left( x \right) \right]\text{d}x}=-x\left[ f\left( x \right)+g\left( x \right) \right]+\int{\left[ f\left( x \right)+g\left( x \right) \right]\text{d}x}\)
\(\Rightarrow -x\left[ f\left( x \right)+g\left( x \right) \right]=C\Rightarrow f\left( x \right)+g\left( x \right)=-\frac{C}{x}\). Vì \(f\left( 1 \right)+g\left( 1 \right)=-C\Rightarrow C=-4\)
Do đó \(f\left( x \right)+g\left( x \right)=\frac{4}{x}\).
Vậy \(I=\int\limits_{1}^{4}{\left[ f\left( x \right)+g\left( x \right) \right]\text{d}x}=8\ln 2\)
