Cho hàm số y = f(x) x^2 khi x lớn hơn hoặc bằng 2 và 2 + x khi x < 2

Dễ thấy, hàm số y = f(x) xác định và liên tục trên $ℝ.$

Ta có: $\sqrt{3x+1}=2\iff 3x+1=4\iff x=1.$

Nhận xét: $3x+1>0\forall x\in \left(0;5\right),$ khi đó $I=\underset{0}{\overset{5}{\int }}\frac{f\left(\sqrt{3x+1}\right)}{\sqrt{3x+1}}dx=\underset{0}{\overset{1}{\int }}\frac{2+\sqrt{3x+1}}{\sqrt{3x+1}}dx+\underset{1}{\overset{5}{\int }}\sqrt{3x+1}dx.$

Xét $I_1=\underset{0}{\overset{1}{\int }}\frac{2+\sqrt{3x+1}}{\sqrt{3x+1}}dx.$ 

Đặt $t=\sqrt{3x+1}\Rightarrow 2tdt=3dx.$

Khi x = 0 thì t = 1, khi x = 1 thì t = 2

Khi đó: $I_1=\underset{1}{\overset{2}{\int }}\left(\frac{2+t}{t}.\frac{2}{3}t\right)dt=\frac{2}{3}\underset{1}{\overset{2}{\int }}\left(2+t\right)dt=\frac{2}{3}\left(2t+\frac{t^2}{2}\right)\left|\begin{array}{l}2\\ 1\end{array}\right.=\frac{2}{3}\left(2.2-2.1+\frac{2^2}{2}-\frac{1}{2}\right)=\frac{7}{3}.$

Xét $I_2=\underset{1}{\overset{5}{\int }}\sqrt{3x+1}dx=\underset{1}{\overset{5}{\int }}{\left(3x+1\right)}^{\frac{1}{2}}\frac{d\left(3x+1\right)}{3}=\frac{1}{3}\left[\frac{{\left(3x+1\right)}^{\frac{3}{2}}}{\frac{3}{2}}\right]\left|\begin{array}{l}5\\ 1\end{array}\right.=\frac{2}{9}\left[\left(3x+1\right)\sqrt{3x+1}\right]\left|\begin{array}{l}5\\ 1\end{array}\right.$

$=\frac{2}{9}\left[\left(3.5+1\right)\sqrt{3.5+1}-\left(3.1+1\right)\sqrt{3.1+1}\right]=\frac{112}{9}.$

Vậy: $I=I_1+I_2=\frac{7}{3}+\frac{112}{9}=\frac{133}{9}.$

Chọn A.