Cho\(I = \int\limits_0^1 {\left( {x + 2} \right)\ln \left( {x + 1} \right){\rm{d}}x}  = a\ln 2 - \frac{7}{b}\) trong đó a, b

* Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l} u = \ln \left( {x + 1} \right)\\ {\rm{d}}v = \left( {x + 2} \right){\rm{d}}x \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {\rm{d}}u = \frac{1}{{x + 1}}{\rm{d}}x\\ v = \frac{1}{2}\left( {{x^2} + 4x + 3} \right) \end{array} \right.\).

Do đó, \(I = \left. {\frac{1}{2}\left( {{x^2} + 4x + 3} \right)\ln \left( {x + 1} \right)} \right|_0^1 - \frac{1}{2}\int\limits_0^1 {\left( {x + 3} \right){\rm{d}}x} \)

\( = \left. {\frac{1}{2}\left( {{x^2} + 4x + 3} \right)\ln \left( {x + 1} \right)} \right|_0^1 - \left. {\frac{1}{4}\left( {{x^2} + 6x} \right)} \right|_0^1 = 4\ln 2 - \frac{7}{4}\)

\( \Rightarrow a = b = 4\)

Vậy \(a + {b^2} = 20\).