Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\mathbb{R}$ thỏa...

Xét $I=\int\limits_{1}^{2}{x{f}'\left( x \right)dx}$. Đặt $\left| f\left( t \right)+m \right|=1\Leftrightarrow \left[ \begin{aligned}
& f\left( t \right)+m=1 \\
& f\left( t \right)+m=-1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& f\left( t \right)=1-m\left( 1 \right) \\
& f\left( t \right)=-1-m\left( 2 \right) \\
\end{aligned} \right.$
Khi đó, ta có
$I=xf\left( x \right)\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.-\int\limits_{1}^{2}{f\left( x \right)dx}=2f\left( 2 \right)-f\left( 1 \right)-\int\limits_{1}^{2}{f\left( x \right)dx}$
$=2f\left( 2 \right)-1-\int\limits_{1}^{2}{f\left( x \right)dx}$
Ta có $f\left( 2x \right)-xf\left( {{x}^{2}} \right)=5x-2{{x}^{3}}-1$.
Thay $x=1\Rightarrow f\left( 2 \right)-f\left( 1 \right)=2\Rightarrow f\left( 2 \right)=3\Rightarrow I=5-\int\limits_{1}^{2}{f\left( x \right)dx}$.
Hơn nữa $f\left( 2x \right)-xf\left( {{x}^{2}} \right)=5x-2{{x}^{3}}-1\Leftrightarrow 2f\left( 2x \right)-2xf\left( {{x}^{2}} \right)=10x-4{{x}^{3}}-2$.
Lấy tích phân 2 vế ta có:
$2\int\limits_{0}^{1}{f\left( 2x \right)dx}-\int\limits_{0}^{1}{2xf\left( {{x}^{2}} \right)dx}=\int\limits_{0}^{1}{\left( 10x-4{{x}^{3}}-2 \right)dx}=2$
$\Leftrightarrow \int\limits_{0}^{1}{f\left( 2x \right)d\left( 2x \right)}-\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)}=2$
$\Leftrightarrow \int\limits_{0}^{2}{f\left( t \right)dt}-\int\limits_{0}^{1}{f\left( u \right)du}=2$
$\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx}=2$
$\Leftrightarrow \int\limits_{1}^{2}{f\left( x \right)dx}=2$
Vậy $I=5-2=3$