Cho hàm số f(x) nhận giá trị khác 0, có đạo hàm liên tục trên R thỏa mãn f(1) = 4

Ta có \({\left( {{x^2} + 3} \right)^2}f'\left( x \right) = 2x \cdot {f^2}\left( x \right)\,\,;\,\,f\left( x \right) \ne 0,\,\,\forall x \in \mathbb{R}.\)

Suy ra \[\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} = \frac{{2x}}{{{{\left( {{x^2} + 3} \right)}^2}}} \Rightarrow \int\limits_1^3 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}} \,{\rm{d}}x = \int\limits_1^3 {\frac{{2x}}{{{{\left( {{x^2} + 3} \right)}^2}}}} \,{\rm{d}}x\]\( \Leftrightarrow \int\limits_1^3 {\frac{{{\rm{d}}\left( {f\left( x \right)} \right)}}{{{f^2}\left( x \right)}}}  = \int\limits_1^3 {\frac{{{\rm{d}}\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}}} \)

\( \Leftrightarrow  - \left. {\frac{1}{{f\left( x \right)}}} \right|_1^3 =  - \left. {\frac{1}{{{x^2} + 3}}} \right|_1^3 \Leftrightarrow \frac{1}{{f\left( 1 \right)}} - \frac{1}{{f\left( 3 \right)}} = \frac{1}{4} - \frac{1}{{12}}\)\( \Leftrightarrow \frac{1}{4} - \frac{1}{{f\left( 3 \right)}} = \frac{1}{4} - \frac{1}{{12}} \Leftrightarrow f\left( 3 \right) = 12.\)

Chú ý công thức: \(\int {\frac{{{\rm{d}}u}}{{{u^2}}}}  =  - \frac{1}{u}.\)

Đáp án: 12.