Biết rằng (x-1)/(x^2+3x+2)dx=aln2+bln3 với a,b thuộc Z

Đáp án D

Phân tích

\(\frac{{x - 1}}{{{x^2} + 3x + 2}} = \frac{{x - 1}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{m}{{x + 1}} + \frac{n}{{x + 2}} \Rightarrow x - 1 = m\left( {x + 2} \right) + n\left( {x + 1} \right).\)

Cho

\(\left\{ \begin{array}{l}x = - 1 \Rightarrow m = - 2\\x = - 2 \Rightarrow n = 3\end{array} \right. \Rightarrow \int\limits_0^1 {\frac{{x - 1}}{{{x^2} + 3x + 2}}} dx = \int\limits_0^1 {\left( {\frac{3}{{x + 2}} - \frac{2}{{x + 1}}} \right)dx} = 3\ln \left| {x + 2} \right|\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle1\atop\scriptstyle}} \right.\)

\( \Rightarrow I = \left( {3\ln 3 - 3\ln 2} \right) - 2\ln 2 = 3\ln 3 - 5\ln 2 \Rightarrow a = - 5,b = 3 \Rightarrow S = - 98\)