Tính giá trị của a+b+c biết \(\int\limits_0^1 {\frac{{xdx}}{{{{\left( {2x + 1} \right)}^2}}} = a + b\ln 2 + c\ln 3} \) với a, b, c là các số hữu tỉ

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}
\int\limits_0^1 {\frac{{xdx}}{{{{\left( {2x + 1} \right)}^2}}} = \int\limits_0^1 {\frac{{\frac{1}{2}\left( {2x + 1} \right) - \frac{1}{2}}}{{{{\left( {2x + 1} \right)}^2}}}} } dx = \frac{1}{2}\int\limits_0^1 {\frac{1}{{2x + 1}}dx - \frac{1}{2}\int\limits_0^1 {\frac{1}{{{{\left( {2x + 1} \right)}^2}}}dx} } \\
 = \left( {\frac{1}{2}.\frac{1}{2}.\ln \left| {2x + 1} \right| - \frac{1}{2}.\frac{1}{2}.\left( { - 1} \right).\frac{1}{{2x + 1}}} \right)\left| \begin{array}{l}
^1\\
_0
\end{array} \right.\\
 = \left( {\frac{1}{4}.\ln \left| {2x + 1} \right| + \frac{1}{4}.\frac{1}{{2x + 1}}} \right)\left| \begin{array}{l}
^1\\
_0
\end{array} \right. = \frac{1}{4}\ln 3 - \frac{1}{6}\\
 \Rightarrow a =  - \frac{1}{6};b = 0;c = \frac{1}{4} \Rightarrow a + b + c = \frac{1}{{12}}
\end{array}\)