Câu hỏi 31

Ta có \({f}'\left( x \right)=\frac{x.f\left( x \right)}{\sqrt{{{x}^{2}}+1}}\Leftrightarrow \frac{{f}'\left( x \right)}{f\left( x \right)}=\frac{x}{\sqrt{{{x}^{2}}+1}}\Leftrightarrow \int{\frac{{f}'\left( x \right)}{f\left( x \right)}\text{d}x}=\int{\frac{x}{\sqrt{{{x}^{2}}+1}}\text{d}x}\)

\(\Leftrightarrow \int{\frac{\text{d}\left( f\left( x \right) \right)}{f\left( x \right)}}=\int{\frac{\text{d}\left( {{x}^{2}}+1 \right)}{2\sqrt{{{x}^{2}}+1}}}=\sqrt{{{x}^{2}}+1}+C\Leftrightarrow \ln f\left( x \right)=\sqrt{{{x}^{2}}+1}+C\Leftrightarrow f\left( x \right)={{e}^{\sqrt{{{x}^{2}}\,+\,1}\,+\,\,C}}\)

Mà \(f\left( 0 \right)=e\) \(\xrightarrow{{}}\) \({{e}^{C\,+\,1}}=e\Rightarrow C=0.\) Vậy \(f\left( \sqrt{3} \right)={{e}^{2}}.\)

Chọn B