Đặt \(5x-3=t\Rightarrow dx=\dfrac{1}{5}dt\) ; \(\left\{{}\begin{matrix}x=-1\Rightarrow t=-8\\x=1\Rightarrow t=2\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{5}\int\limits^2_{-8}f\left(\left|t\right|\right)dt=\dfrac{1}{5}\int\limits^0_{-8}f\left(-t\right)dt+\dfrac{1}{5}\int\limits^2_0f\left(t\right)dt=\dfrac{1}{5}.4+\dfrac{1}{5}.2=\dfrac{6}{5}\)