Cho hàm số $y=f\left( x \right)$ liên tục trên đoạn $\left[ -3;1...

Đặt $t={{x}^{2}}-3\Rightarrow 2x\text{d}x=\text{d}t$.
Suy ra
$I=\int_{0}^{2}{({{x}^{3}}+x)}{f}'({{x}^{2}}-3)\ \text{d}x$ $=\dfrac{1}{2}\int_{0}^{2}{2}x({{x}^{2}}-3+4){f}'({{x}^{2}}-3)\ \text{d}x$ $=\dfrac{1}{2}\int_{-3}^{1}{(t+4)}{f}'(t)\ \text{d}t$ $\Rightarrow 2I=\int_{-3}^{1}{(x+4)}{f}'(x)\ \text{d}x$.
Đặt $\left\{ \begin{aligned}
& u=x+4 \\
& \text{d}v=f'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.$.
Ta có $2I=\int_{-3}^{1}{(x+4)}{f}'(x)\ \text{d}x=(x+4)f(x)|_{-3}^{1}-\int_{-3}^{1}{f}(x)\ \text{d}x=-\int_{-3}^{1}{f}(x)\ \text{d}x$
$=-\int_{-3}^{-1}{f}(x)\ \text{d}x-\int_{-1}^{0}{f}(x)\ \text{d}x-\int_{0}^{1}{f}(x)\ \text{d}x$ $=-27+2-3=-28\Rightarrow I=-14.$