Cho tích phân \(\int\limits_2^3 {\frac{1}{{{x^3} + {x^2}}}dx = a\ln 3 + b\ln 2 + c} \) với \(a,b,c \in \mathbb{Q}\). Tính \(S = a +

Chung Thanh

New member
Cho tích phân \(\int\limits_2^3 {\frac{1}{{{x^3} + {x^2}}}dx = a\ln 3 + b\ln 2 + c} \) với \(a,b,c \in \mathbb{Q}\). Tính \(S = a + b + c.\)
A. \(S = - \frac{2}{3}\)
B. \(S = - \frac{7}{6}\)
C. \(S = \frac{2}{3}\)
D. \(S = \frac{7}{6}\)
 
Ta có \(\int\limits_2^3 {\frac{1}{{{x^3} + {x^2}}}} dx = \int\limits_2^3 {\left( {\frac{1}{{{x^2}}} - \frac{1}{x} + \frac{1}{{x + 1}}} \right)} dx = \left( { - \frac{1}{x} - \ln \left| x \right| + \ln \left| {x + 1} \right|} \right)\left| {\begin{array}{*{20}{c}}3\\2\end{array}} \right. = - 2\ln 3 + 3\ln 2 + \frac{1}{6}\)
Suy ra \(\left\{ {\begin{array}{*{20}{c}}{a = - 2,b = 3}\\{c = \frac{1}{6}}\end{array}} \right. \Rightarrow S = \frac{7}{6}.\)