Cho hàm số y = f(x) có đạo hàm liên tục trên đoạn 0;1 và thỏa mãn

Đáp án C

Ta có: $f^2\left(x\right)=xf\left(x\right)f\text{'}\left(x\right)+2x+4$

$\Rightarrow I=\underset{0}{\overset{1}{\int }}{f}^2\left(x\right)dx=\underset{0}{\overset{1}{\int }}\left[xf\left(x\right)f\text{'}\left(x\right)+2x+4\right]dx=\underset{0}{\overset{1}{\int }}xf\left(x\right)f\text{'}\left(x\right)dx+\underset{0}{\overset{1}{\int }}\left(2x+4\right)dx=A+5\left(*\right)$

Tính $A=\underset{0}{\overset{1}{\int }}xf\left(x\right)f\text{'}\left(x\right)dx$

Đặt $\left\{\begin{array}{l}u=xf\left(x\right)\\ dv=f\text{'}\left(x\right)dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=\left(f\left(x\right)+xf\text{'}\left(x\right)\right)dx\\ v=f\left(x\right)\end{array}\right.$

$\Rightarrow A=x{f}^2{\left.\left(x\right)\right|}_0^1-\underset{0}{\overset{1}{\int }}f\left(x\right)\left(f\left(x\right)+xf\text{'}\left(x\right)\right)dx=9-\underset{0}{\overset{1}{\int }}{f}^2\left(x\right)dx-\underset{0}{\overset{1}{\int }}xf\left(x\right)f\text{'}\left(x\right)dx$

$\Rightarrow A=\frac{9-I}{2}\left(2*\right)$. Thay (2*) vào (*), ta được: $I=\frac{9-I}{2}+5\iff I=\frac{19}{3}$