Cho hàm số $y=f\left( x \right)$ liên tục, có đạo hàm trên đoạn...

Ta có: $f\left( x \right)+2xf\left( {{x}^{2}} \right)+3{{x}^{2}}f\left( {{x}^{3}} \right)=\sqrt{1-{{x}^{2}}}\ \forall x\in \left[ 0;1 \right]$.
Lấy tích phân 2 vế cận từ 0 đến 1 ta có: $\int\limits_{0}^{1}{\left[ f\left( x \right)+2xf\left( {{x}^{2}} \right)+3{{x}^{2}}f\left( {{x}^{3}} \right) \right]dx}=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}dx}$.
Ta có: $VT=\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)}+\int\limits_{0}^{1}{f\left( {{x}^{3}} \right)d\left( {{x}^{3}} \right)}$.
Mặt khác: $B=\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)}\xrightarrow{t={{x}^{2}}}B=\int\limits_{0}^{1}{f\left( t \right)dt}=\int\limits_{0}^{1}{f\left( x \right)dx}$.
Tương tự ta có: $C=\int\limits_{0}^{1}{f\left( {{x}^{3}} \right)d\left( {{x}^{3}} \right)}=\int\limits_{0}^{1}{f\left( x \right)dx\Rightarrow VT=3\int\limits_{0}^{1}{f\left( x \right)dx}}$.
Lại có: $VP=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}dx}$. Đặt $x=\sin u\Rightarrow dx=\cos udu$, đổi cận $\left| \begin{aligned}
& x=0\Rightarrow u=0 \\
& x=1\Rightarrow u=\dfrac{\pi }{2} \\
\end{aligned} \right.$
Khi đó $\begin{aligned}
& VP=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{1-{{\sin }^{2}}u}\cos udu}=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}udu}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2u \right)du} \\
& \ \ \ \ \ =\dfrac{1}{2}\left( 1+\dfrac{\sin 2u}{2} \right)\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{\pi }{4}\Rightarrow 3\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{12} \\
\end{aligned}$