Tích phân \(I = \int\limits_1^2 {\frac{{ax + 1}}{{...

D

- Hướng dẫn giải

Ta có:

\(I = \int\limits_1^2 {\frac{{ax + 1}}{{{x^2} + 3x + 2}}} dx \\= a\int\limits_1^2 {\frac{x}{{{x^2} + 3x + 2}}} dx + \int\limits_1^2 {\frac{1}{{{x^2} + 3x + 2}}} dx\).

Xét

\({I_1} = a\int\limits_1^2 {\frac{x}{{{x^2} + 3x + 2}}} dx \\= a\int\limits_1^2 {\left( {\frac{2}{{x + 2}} - \frac{1}{{x + 1}}} \right)} dx \\= a\left. {\left( {2\ln \left| {x + 2} \right| - \ln \left| {x + 1} \right|} \right)} \right|_1^2 \\= a\left( {2\ln 4 - 3\ln 3 + \ln 2} \right) \\= 2a\ln \frac{4}{3} + a\ln \frac{2}{3}\)

Xét \({I_2} = \int\limits_1^2 {\frac{1}{{{x^2} + 3x + 2}}} dx = \left. {\left( {\ln \left| {x + 1} \right| - \ln \left| {x + 2} \right|} \right)} \right|_1^2 = - \ln \frac{4}{3} - \ln \frac{2}{3}\).

\( \Rightarrow I = {I_1} + I{}_2 = \left( {2a - 1} \right)\ln \frac{4}{3} + \left( {a - 1} \right)\ln \frac{2}{3}\)

Theo đề bài: \(I = \frac{3}{5}\ln \frac{4}{3} + \frac{3}{5}\ln \frac{2}{3} \Rightarrow a = \frac{4}{5}\).