\(\begin{array}{l} \int\limits_0^2 {\frac{{5x + 7}}{{{x^3} + 3x + 2}}dx} \\ = \int\limits_0^2 {\frac{{5(2x + 3)}}{{2({x^2} + 3x + 2)}}} dx - \int\limits_0^2 {\frac{1}{{2({x^2} + 3x + 2)}}} dx \end{array}\)
Tính \({I_1} = \int\limits_0^2 {\frac{{5(2x + 3)}}{{2({x^2} + 3x + 2)}}} dx\)
Đặt \(u = {x^2} + 3x + 2 \Rightarrow du = \left( {2x + 3} \right)dx\)
Khi đó: \({I_1} = \frac{5}{2}\int\limits_2^{12} {\frac{1}{u}} = \left. {\frac{5}{2}\ln \left| u \right|} \right|_2^{12} = \frac{5}{2}\ln 12 - \frac{5}{2}\ln 2\)
Tính \({I_2} = \int\limits_0^2 {\frac{1}{{2({x^2} + 3x + 2)}}} dx\)
\(\begin{array}{l} {I_2} = \int\limits_0^2 {\frac{{(x + 2) - (x + 1)}}{{2(x + 2)(x + 1)}}dx} \\ = \int\limits_0^2 {\frac{1}{{2(x + 1)}}dx} - \int\limits_0^2 {\frac{1}{{2(x + 2)}}dx} \\ = \left. {\frac{1}{2}\left[ {\ln \left| {x + 1} \right| - \ln \left| {x + 2} \right|} \right]} \right|_0^2\\ = \frac{1}{2}\ln 3 - \frac{1}{2}\ln 4 + \frac{1}{2}\ln 2 \end{array}\)
Vậy:
\(\begin{array}{l} I = {I_1} - {I_2} = \frac{5}{2}\ln 12 - \frac{5}{2}\ln 2 - \frac{1}{2}\ln 3 + \frac{1}{2}\ln 4 + \frac{1}{2}\ln 2\\ = 2\ln 3 + 3\ln 4 - 3\ln 2 = 2\ln 3 + 3\ln 2 \end{array}\)
