Giả sử hàm f có đạo hàm cấp hai trên $\mathbb{R}$ thỏa mãn...

Từ giả thiết $f\left( 1-x \right)+{{x}^{2}}{f}''\left( x \right)=2x\Rightarrow f\left( 1 \right)=0$ (thay $x=0$ )
$\Rightarrow \int\limits_{0}^{1}{{{x}^{2}}{f}''\left( x \right)dx}=\int\limits_{0}^{1}{2xdx}-\int\limits_{0}^{1}{f\left( 1-x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={f}''\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v={f}'\left( x \right) \\
\end{aligned} \right.$
Khi đó $\int\limits_{0}^{1}{{{x}^{2}}{f}''\left( x \right)dx}={{x}^{2}}{f}'\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-2\int\limits_{0}^{1}{x{f}'\left( x \right)dx}=1-2I$
Mặt khác $\int\limits_{0}^{1}{2xdx}-\int\limits_{0}^{1}{f\left( 1-x \right)dx}={{x}^{2}}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right)dx}=1-\int\limits_{0}^{1}{f\left( x \right)dx}=1-xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.+\int\limits_{0}^{1}{x{f}'\left( x \right)dx}=1+I$
Suy ra $1-2I=1+I\Rightarrow I=0$.