Ta có: [tex]\int \frac{f(\sqrt{x+1})}{\sqrt{x+1}}dx=2\int f(\sqrt{x+1})d(\sqrt{x+1})=2.F(\sqrt{x+1})+C[/tex]
Do đó : [tex]2.F(\sqrt{x+1})+C=\frac{2.(\sqrt{x+1}+3)}{x+5}+C=\frac{2.(\sqrt{x+1}+3)}{\sqrt{x+1}^2+4}+C[/tex]
=> F(x)=[tex]\frac{x+3}{x^2+4}[/tex]
Vậy F(2x)=[tex]\frac{2x+3}{4x^2+4}[/tex] ....chọn C