Giả sử \(\int\limits_{1}^{2}{\left( 2x-1 \right)\ln x\text{d}x}=a\ln 2+b\), \(\left( a;b\in \mathbb{Q} \right)\). Tính \(a+b\).

* Hướng dẫn giải

\(\left\{ \begin{array}{l} u = \ln x\\ dv = \left( {2x - 1} \right)dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = {x^2} - x \end{array} \right.\)

\(\int\limits_{1}^{2}{\left( 2x-1 \right)\ln x}\text{d}x=\)\(=\left. \left[ \left( {{x}^{2}}-x \right)\ln x \right] \right|_{1}^{2}-\int\limits_{1}^{2}{\frac{{{x}^{2}}-x}{x}}\text{d}x\)\(=2\ln 2-\left. \left( \frac{{{x}^{2}}}{2}-x \right) \right|_{1}^{2}\)\(=2\ln 2-\frac{1}{2}\) nên \(a=2\), \(b=-\frac{1}{2}\).

Vậy \(a+b\)\)=\frac{3}{2}\).