Đặt \(\sqrt{2x+1}+2=t\Rightarrow2x=t^2-4t+3\Rightarrow dx=\left(t-2\right)dt\)
\(\left\{{}\begin{matrix}x=0\Rightarrow t=3\\x=4\Rightarrow t=5\end{matrix}\right.\)
\(E=\int\limits^5_3\dfrac{2\left(t^2-4t+3\right)-1}{t}.\left(t-2\right)dt=\int\limits^5_3\left(2t^2-12t+21-\dfrac{10}{t}\right)dt\)
\(=\left(\dfrac{2t^3}{3}-6t^2+21t-10ln\left|t\right|\right)|^5_3=\dfrac{34}{3}-10ln\left(\dfrac{5}{3}\right)\)
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