$\displaystyle \begin{array}{{>{\displaystyle}l}} [ f'( x)]^{2} =( x+1) f( x)\\ Do\ f( x) >0,\ nên\ ta\ có:\\ \frac{[ f'( x)]^{2}}{f( x)} =x+1\ \Leftrightarrow \frac{f'( x)}{\sqrt{f( x)}} =\sqrt{x+1} ,\ lấy\ tích\ phân\ 2\ vế\ ta\ được:\\ \int \frac{f'( x)}{\sqrt{f( x)}} dx=\int ( x+1) dx\\ \Leftrightarrow \int f( x)^{\frac{1}{2}} df( x) =\int ( x+1)^{\frac{1}{2}} d( x+1)\\ \Leftrightarrow 2\sqrt{f( x)} =\frac{1}{2\sqrt{x+1}} +C\\ Có\ 2\sqrt{f( 3)} =\frac{1}{2\sqrt{3+1}} +C\Leftrightarrow 2\sqrt{\frac{3}{2}} -\frac{1}{4} =\frac{-1+4\sqrt{6}}{4} =C\\ Vâỵ\ \sqrt{f( x)} =\frac{1}{4\sqrt{x+1}} +\frac{-1+4\sqrt{6}}{8}\\ \end{array}$
