Biết rằng (x^3+2)/(x^2+x)dx=a+bln2+cln3+dln5

Đáp án A

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

\( \Rightarrow x + 2 = m\left( {x + 1} \right) + nx,\) cho \(\left\{ \begin{array}{l}x = 0 \Rightarrow m = 2\\x = - 1 \Rightarrow n = - 1\end{array} \right. \Rightarrow \int\limits_2^4 {\frac{{{x^3} + 2}}{{{x^2} + x}}dx} = \int\limits_2^4 {\left( {x - 1 + \frac{2}{x} - \frac{1}{{x + 1}}} \right)dx} \)

\( \Rightarrow I = \left. {\left( {\frac{{{x^2}}}{2} - x + 2\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)} \right|_2^4 = \left( {4 + 2\ln 4 - \ln 5} \right) - \left( {2\ln 2 - \ln 3} \right)\)

\( = 4 + 2\ln 2 + \ln 3 - \ln 5 \Rightarrow a = 4,{\rm{ }}b = 2,{\rm{ }}c = 1,{\rm{ }}d = - 1 \Rightarrow S = 6.\)