Cho \(\int\limits_0^1 {\frac{{x{\rm{d}}x}}{{{{\left( {x + 2} \right)}^2}}}}  = a + b\ln 2 + c\ln 3\) với \(a, b, c\) là các số hữu t

* Hướng dẫn giải

\(\int\limits_0^1 {\frac{{x{\rm{d}}x}}{{{{\left( {x + 2} \right)}^2}}}}  = \int\limits_0^1 {\frac{{\left( {x + 2} \right) - 2}}{{{{\left( {x + 2} \right)}^2}}}{\rm{d}}x = \int\limits_0^1 {\frac{{{\rm{d}}x}}{{x + 2}}} }  - \int\limits_0^1 {\frac{{2{\rm{d}}x}}{{{{\left( {x + 2} \right)}^2}}}} \)

\( = \left. {\ln \left( {x + 2} \right)} \right|_0^1 - \left. {2.\frac{{{{\left( {x + 2} \right)}^{ - 1}}}}{{ - 1}}} \right|_0^1 = \ln 3 - \ln 2 + \frac{2}{3} - 1 =  - \frac{1}{3} - \ln 2 + \ln 3\)

Vậy \(a =  - \frac{1}{3};b =  - 1;c = 1 \Rightarrow 3a + b + c =  - 1\)