Giả sử intlimits_1^2 {(2x - 1)ln xdx = aln 2 + b,(a,b in mathbb{Q}). Tính tổng S=a+b

Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = \ln x}\\ {dv = (2x - 1)dx} \end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {du = \frac{{dx}}{x}}\\ {v = {x^2} - x} \end{array}} \right.} \right.\)

\(\Rightarrow I = \int\limits_1^2 {(2x - 1)\ln xdx = } \left. {({x^2} - x)\ln x} \right|_1^2 - \int\limits_1^2 {(x - 1)dx}\)

\(\Leftrightarrow I = \left. {({x^2} - x)\ln x} \right|_1^2 - \left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|_1^2 = 2\ln 2 - \frac{1}{2} \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a = 2}\\ {b = - \frac{1}{2}} \end{array}} \right.\)

\(\Rightarrow a + b = \frac{3}{2}\)