Cho hàm số f(x) có đạo hàm liên tục trên [-1;1] và thỏa mãn f(1) = 0,

Ta có:

${\left(f^{\text{'}}\left(x\right)\right)}^2+4f\left(x\right)=8{\text{x}}^2+16\text{x}-8\Rightarrow \underset{-1}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)\right]}^{2}d\text{x}+2\underset{-1}{\overset{1}{\int }}2f\left(x\right)d\text{x}=\underset{-1}{\overset{1}{\int }}\left(8{\text{x}}^2+16\text{x}-8\right)d\text{x}$ (1).

Xét $I=\underset{-1}{\overset{1}{\int }}2f\left(x\right)d\text{x}$, đặt $\left\{\begin{array}{l}u=f\left(x\right)\\ dv=2\text{dx}\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=f^{\text{'}}\left(x\right)d\text{x}\\ v=2\text{x}+2\end{array}\right.$.

Do đó $I=\underset{-1}{\overset{1}{\int }}2f\left(x\right)d\text{x}={\left.\left(2\text{x}+2\right)f\left(x\right)\right|}_{-1}^1-\underset{-1}{\overset{1}{\int }}\left(2\text{x}+2\right)f^{\text{'}}\left(x\right)d\text{x}=-\underset{-1}{\overset{1}{\int }}\left(2x+2\right)f^{\text{'}}\left(x\right)d\text{x}$.

Từ (1) suy ra $\underset{-1}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)\right]}^{2}d\text{x}+2\underset{-1}{\overset{1}{\int }}2f\left(x\right)d\text{x}=\underset{-1}{\overset{1}{\int }}\left(8{\text{x}}^2+16\text{x}-8\right)d\text{x}$

$\iff \underset{-1}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)\right]}^{2}d\text{x}-2\underset{-1}{\overset{1}{\int }}\left(2\text{x}+2\right)f^{\text{'}}\left(x\right)d\text{x}+\underset{-1}{\overset{1}{\int }}{\left(2\text{x}+2\right)}^{2}d\text{x}=\underset{-1}{\overset{1}{\int }}\left(12{\text{x}}^2+24\text{x}-4\right)d\text{x}$

$\iff \underset{-1}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)-\left(2\text{x}+2\right)\right]}^{2}d\text{x}=0\Rightarrow f^{\text{'}}\left(x\right)=2\text{x}+2\Rightarrow f\left(x\right)=x^2+2\text{x}+C$.