Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[...

Đặt $t=\sqrt{x}\Leftrightarrow {{t}^{2}}=x\Leftrightarrow dx=2tdt$ và $\left\{ \begin{array}{*{35}{l}}
x=0\Rightarrow t=0 \\
x=1\Rightarrow t=1 \\
\end{array} \right..$
Khi đó $\underset{0}{\overset{1}{\mathop \int }} f\left( \sqrt{x} \right)dx=\underset{0}{\overset{1}{\mathop \int }} 2t.f\left( t \right)dt=2\underset{0}{\overset{1}{\mathop \int }} x.f\left( x \right)dx=\dfrac{2}{5}\Leftrightarrow \underset{0}{\overset{1}{\mathop \int }} x.f\left( x \right)dx=\dfrac{1}{5}.$
Đặt $\left\{ \begin{array}{*{35}{l}}
u=f\left( x \right) \\
dv=xdx \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
du={f}'\left( x \right)dx \\
v=\dfrac{{{x}^{2}}}{2} \\
\end{array} \right.$ (từng phần)
$\Rightarrow \int\limits_{0}^{1}{x.f\left( x \right)dx}=\left. \dfrac{{{x}^{2}}.f\left( x \right)}{2} \right|_{0}^{1}-\dfrac{1}{2}\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx\Rightarrow \int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}=\dfrac{3}{5}}.$
Xét $\underset{0}{\overset{1}{\mathop \int }} {{\left[ {f}'\left( x \right)+k{{x}^{2}} \right]}^{2}}dx=\underset{0}{\overset{1}{\mathop \int }} {{\left[ {f}'\left( x \right) \right]}^{2}}dx+2k\underset{0}{\overset{1}{\mathop \int }} {{x}^{2}}{f}'\left( x \right)dx+{{k}^{2}}\underset{0}{\overset{1}{\mathop \int }} {{x}^{4}}dx$
$=\dfrac{9}{5}+\dfrac{6}{5}k+\dfrac{1}{5}{{k}^{2}}=0\Leftrightarrow \dfrac{1}{5}{{\left( k+3 \right)}^{2}}=0\Leftrightarrow k=-3.$
Do đó ${f}'\left( x \right)-3{{x}^{2}}=0\Leftrightarrow {f}'\left( x \right)=3{{x}^{2}}\Leftrightarrow f\left( x \right)=\mathop{\int }^{}{f}'\left( x \right)dx={{x}^{3}}+C$
Mà $f\left( 1 \right)=1\Rightarrow C=0$ nên $f\left( x \right)={{x}^{3}}\to I=\int\limits_{0}^{1}{{{x}^{3}}dx}=\left. \dfrac{{{x}^{4}}}{4} \right|_{0}^{1}=\dfrac{1}{4}.$ Chọn C