Đặt \(\sqrt{8x+1}=t\Rightarrow8x+1=t^2\Rightarrow8dx=2tdt\Rightarrow dx=\frac{1}{4}tdt\)
\(\left\{{}\begin{matrix}x=1\Rightarrow t=3\\x=3\Rightarrow t=5\end{matrix}\right.\)
\(\Rightarrow I=\frac{1}{4}\int\limits^5_3\frac{t}{1+t}dt=\frac{1}{4}\int\limits^5_3\left(1-\frac{1}{1+t}\right)dt=\frac{1}{4}\left(t-ln\left(1+t\right)\right)|^5_3=\frac{1}{2}-\frac{1}{4}ln3+\frac{1}{4}ln2\)
\(\Rightarrow a+b+c=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
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