Biết \(\int\limits_1^2 {\frac{{{x^2} + x + 1}}{{x + 1}}dx = a + \ln b} \). Khi đó \(a + b\) bằng.

* Hướng dẫn giải

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

\( \Rightarrow a = b = \frac{3}{2}\)\( \Rightarrow a + b = \frac{3}{2} + \frac{3}{2} = 3.\)