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:
$\sqrt{1+{{f}^{2}}\left( x \right)}-3={{x}^{2}}+x\Rightarrow \sqrt{1+{{f}^{2}}\left( x \right)}-3=1+1\Leftrightarrow \sqrt{1+{{f}^{2}}\left( x \right)}=5$
$\Leftrightarrow 1+{{f}^{2}}\left( 1 \right)=25\Leftrightarrow {{f}^{2}}\left( 1 \right)=24\Leftrightarrow f\left( 1 \right)=\sqrt{24}$.