Ta có: ${\left[{\text{f}}^{\text{'}}\left(\text{x}\right)\right]}^2=4\cdot \left[2{\text{x}}^2+1-\text{f}\left(\text{x}\right)\right]$

$$\begin{gathered} \iff {\left[f^{\text{'}}\left(x\right)\right]}^2-4x.f^{\text{'}}\left(x\right)+4x^2=12x^2+4-4\left[x.f^{\text{'}}\left(x\right)+f\left(x\right)\right] \\ \iff {\left[f^{\text{'}}\left(x\right)-2x\right]}^2=12x^2+4-4{[x.f(x\left)\right]}^{\text{'}} \\ \Rightarrow {\int }_0^1{\left[f^{\text{'}}\left(x\right)-2x\right]}^{2}dx={\int }_0^1\left(12x^2+4\right)dx-4{\int }_0^1{\left[x.f\left(x\right)\right]}^{\text{'}}dx \\ \iff {\int }_0^1{\left[f^{\text{'}}\left(x\right)-2x\right]}^{2}dx=8-{\left.4\left(x.f\left(x\right)\right)\right|}_0^1 \\ \iff {\int }_0^1{\left[f^{\text{'}}\left(x\right)-2x\right]}^{2}dx=8-4.f\left(1\right) \\ \iff {\int }_0^1{\left[f^{\text{'}}\left(x\right)-2x\right]}^{2}dx=0 \\ \Rightarrow f^{\text{'}}\left(x\right)=2x. \end{gathered}$$

Từ đó: $f\left(x\right)=\int f^{\text{'}}\left(x\right)dx=\int 2xdx=x^2+C$ mà $f\left(1\right)=2\Rightarrow C=1$ nên $f\left(x\right)=x^2+1$.