Cho hàm số $y=f\left( x \right)$ có ${f}'\left( x...

Với mọi $x\in \left[ 1;4 \right]$, ta có $x+2xf\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow {f}'\left( x \right)=\sqrt{x\left( 1+2f\left( x \right) \right)}\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}=\sqrt{x}$
$\Rightarrow \int{\dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}}dx=\int{\sqrt{x}dx,\left( * \right).}$
Đặt $t=\sqrt{1+2f\left( x \right)}\Rightarrow 2tdt=2{f}'\left( x \right)dx$ thì $\int{\dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}}dx=\int{\dfrac{tdt}{t}=t+{{C}_{1}}}=\sqrt{1=2f\left( x \right)}+{{C}_{2}}$
Do đó $\left( * \right)\Leftrightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}x\sqrt{x}+C.$ Vì $f\left( 1 \right)=\dfrac{3}{2}\Rightarrow C=\dfrac{4}{3}.$
Suy ra $f\left( x \right)=\dfrac{1}{2}{{\left( \dfrac{2}{3}x\sqrt{x}=\dfrac{4}{3} \right)}^{2}}-\dfrac{1}{2}.$ Vậy $f\left( 4 \right)=\dfrac{391}{18}.$