Biết rằng intlimits_1^5 {frac{3}{{{x^2} + 3x}}dx} = aln 5 + bln 2, left( {a,b in Z } ight)

* Hướng dẫn giải

\(\begin{array}{l} \int\limits_1^5 {\frac{3}{{{x^2} + 3x}}dx} = \int\limits_1^5 {\frac{3}{{x(x + 3)}}dx} = \int\limits_1^5 {\left( {\frac{1}{x} - \frac{1}{{x + 3}}} \right)dx} \\ = \left. {\left( {\ln \left| x \right|} \right)} \right|_1^5 - \left. {\left( {\ln \left| {x + 3} \right|} \right)} \right|_1^5 = \ln 5 - \ln 2\\ \Rightarrow a + b = 0. \end{array}\)