Biết \(I=\int\limits_{1}^{5}{\frac{2\left| x-2 \right|+1}{x}}\text{d}x=4+a\ln 2+b\ln 5\) với \(a,b\in \mathbb{Z}\). Tính S=a+b.

* Hướng dẫn giải

Ta có \(\left| x-2 \right|=\left\{ \begin{align} & x-2.\text{ Khi }x\ge 2 \\ & 2-x.\text{ Khi }x\le 2 \\ \end{align} \right.\).

Do đó \(I=\int\limits_{1}^{2}{\frac{2\left| x-2 \right|+1}{x}}\text{ d}x+\int\limits_{2}^{5}{\frac{2\left| x-2 \right|+1}{x}\text{ }}\text{d}x\).

\(=\int\limits_{1}^{2}{\frac{2\left( 2-x \right)+1}{x}}\text{ d}x+\int\limits_{2}^{5}{\frac{2\left( x-2 \right)+1}{x}\text{ }}\text{d}x\).

\(=\int\limits_{1}^{2}{\left( \frac{5}{x}-2 \right)}\text{ d}x+\int\limits_{2}^{5}{\left( 2-\frac{3}{x} \right)\text{ }}\text{d}x\).

\(=\left( 5\ln x-2x \right)\left| \begin{matrix} 2 \\ 1 \\ \end{matrix}+ \right.\left( 2x-5\ln x \right)\left| \begin{matrix} 5 \\ 2 \\ \end{matrix} \right.\).

\(=4+8\ln 2-3\ln 5\).

⇒ \(\left\{ \begin{align} & a=8 \\ & b=-3 \\ \end{align} \right.\) ⇒ S=a+b=5.