\(I=\int\limits^2_1\frac{ln\left(1+2x\right)}{x^2}dx\)
Đặt \(\left\{{}\begin{matrix}u=ln\left(1+2x\right)\\dv=\frac{dx}{x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{2}{1+2x}dx\\v=-\frac{1}{x}\end{matrix}\right.\)
\(\Rightarrow I=-\frac{1}{x}.ln\left(1+2x\right)|^2_1+\int\limits^2_1\frac{2dx}{x\left(2x+1\right)}=-\frac{1}{2}ln5+ln3+I_1\)
\(I_1=\int\limits^2_1\frac{4dx}{2x\left(2x+1\right)}=4\int\limits^2_1\left(\frac{1}{2x}-\frac{1}{2x+1}\right)dx=2ln\left(\frac{2x}{2x+1}\right)|^2_1=2ln2+2ln3-2ln5\)
\(\Rightarrow I=-\frac{5}{2}ln5+3ln3+2ln2\) \(\Rightarrow\left\{{}\begin{matrix}a=-5\\b=3\\c=2\end{matrix}\right.\) \(\Rightarrow a+2\left(b+c\right)=5\)
