Cho \[\mathop \smallint \limits_0^1 \left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right)dx = a\ln 2 + b\ln 3\]với a, b là các số nguyên. Chứng minh a + 2b = 0.
\[\mathop \smallint \limits_0^1 \left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right)dx = \left( {ln|x + 1| - ln|x + 2|} \right)|_0^1\]
\[ = ln\mid \frac{{x + 1}}{{x + 2}}\mid \mid _0^1 = ln\frac{2}{3} - ln\frac{1}{2} = ln2 - ln3 + ln2 = 2ln2 - ln3\]
\[ \Rightarrow \left\{ \begin{array}{l}a = 2\\b = - 1\end{array} \right. \Rightarrow a + 2b = 2 - 2 = 0\]
Vậy a + 2b = 0.