Cho hàm số f(x) liên tục trên \(\mathbb{R}\) thỏa mãn các điều kiện: f\[\left( 0 \right) = 2\sqrt 2 ,\;f(x) > 0,\forall x \in \mathbb{R}\;\] và \[f(x).f\prime (x) = (2x + 1)\sqrt {1 + {f^2}(x)} ,\forall x \in \mathbb{R}\]. Khi đó giá trị f(1) bằng

Ta có:\[f\left( x \right).f'\left( x \right) = \left( {2x + 1} \right)\sqrt {1 + {f^2}\left( x \right)} \]

\[ \Rightarrow \frac{{f\left( x \right).f'\left( x \right)}}{{\sqrt {1 + {f^2}\left( x \right)} }} = 2x + 1 \Rightarrow \smallint \frac{{f\left( x \right).f'\left( x \right)}}{{\sqrt {1 + {f^2}\left( x \right)} }}dx = \smallint \left( {2x + 1} \right)dx\]

Tính\[\smallint \frac{{f\left( x \right).f'\left( x \right)}}{{\sqrt {1 + {f^2}\left( x \right)} }}dx\]  ta đặt

\[\sqrt {1 + {f^2}\left( x \right)} = t \Rightarrow 1 + {f^2}\left( x \right) = {t^2} \Rightarrow 2f\left( x \right)f'\left( x \right)dx = 2tdt\]

\[ \Rightarrow f\left( x \right)f'\left( x \right)dx = tdt\]

Thay vào ta được

\[\smallint \frac{{f\left( x \right).f'\left( x \right)}}{{\sqrt {1 + {f^2}\left( x \right)} }}dx = \smallint \frac{{tdt}}{t} = \smallint dt = t + C = \sqrt {1 + {f^2}\left( x \right)} + C\]

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:

\[\begin{array}{*{20}{l}}{\sqrt {1 + {f^2}\left( x \right)} - 3 = {x^2} + x \Rightarrow \sqrt {1 + {f^2}\left( 1 \right)} - 3 = 1 + 1 \Leftrightarrow \sqrt {1 + {f^2}\left( 1 \right)} = 5}\\{ \Leftrightarrow 1 + {f^2}\left( 1 \right) = 25 \Leftrightarrow {f^2}\left( 1 \right) = 24 \Leftrightarrow f\left( 1 \right) = \sqrt {24} }\end{array}\]