Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn...

Phương pháp:
Chia cả hai vế cho $\sqrt{1+{{f}^{2}}\left( x \right)}$ rồi lấy nguyên hàm hai vế tìm $f\left( x \right)$.
Cách giải:
Ta có $f\left( x \right).f'\left( x \right)=2x+1\sqrt{1+{{f}^{2}}\left( x \right)}$.
$\Rightarrow \dfrac{f\left( x \right).f'\left( x \right)}{\sqrt{1+{{f}^{2}}\left( x \right)}}=2x+1\Rightarrow \int{\dfrac{f\left( x \right).f'\left( x \right)}{\sqrt{1+{{f}^{2}}\left( x \right)}}dx}=\int{\left( 2x+1 \right)dx}$
Tính $\int{\dfrac{f\left( x \right).f'\left( x \right)}{\sqrt{1+{{f}^{2}}\left( x \right)}}dx}$, ta đặt $\sqrt{1+{{f}^{2}}\left( x \right)}=t\Rightarrow 1+{{f}^{2}}\left( x \right)={{t}^{2}}\Leftrightarrow 2f\left( x \right)f'\left( x \right)dx=2tdt\Rightarrow f\left( x \right).f'\left( x \right)dx=tdt$.
Thay vào ta được $\int{\dfrac{f\left( x \right).f'\left( x \right)}{\sqrt{1+{{f}^{2}}\left( x \right)}}dx}=\int{\dfrac{tdt}{t}}=\int{dt}=t+C=\sqrt{1+{{f}^{2}}\left( x \right)}+C$.
Do đó $\sqrt{1+{{f}^{2}}\left( x \right)}+C={{x}^{2}}+x$.
$f\left( 0 \right)=2\sqrt{2}\Rightarrow \sqrt{1+{{\left( 2\sqrt{2} \right)}^{2}}}+C=0\Leftrightarrow C=-3$.
Từ đó: $\begin{aligned}
& \sqrt{1+{{f}^{2}}\left( x \right)}-3={{x}^{2}}+x\Rightarrow \sqrt{1+{{f}^{2}}\left( x \right)}-3=1+1\Leftrightarrow \sqrt{1+{{f}^{2}}\left( x \right)}=5 \\
& \Leftrightarrow 1+{{f}^{2}}\left( 1 \right)=25\Leftrightarrow {{f}^{2}}\left( 1 \right)=24\Leftrightarrow f\left( 1 \right)=\sqrt{24} \\
\end{aligned}$