Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\int\limits_3^4 {\frac{1}{{{x^2}\left( {x + 2} \right)}}dx} \\
= \frac{1}{2}.\int\limits_3^4 {\frac{2}{{{x^2}\left( {x + 2} \right)}}dx} \\
= \frac{1}{2}.\int\limits_3^4 {\frac{{\left( {x + 2} \right) - x}}{{{x^2}\left( {x + 2} \right)}}dx} \\
= \frac{1}{2}\int\limits_3^4 {\left[ {\frac{1}{{{x^2}}} - \frac{1}{{x\left( {x + 2} \right)}}} \right]dx} \\
= \frac{1}{2}.\int\limits_3^4 {\left[ {\frac{1}{{{x^2}}} - \frac{1}{2}.\frac{{\left( {x + 2} \right) - x}}{{x\left( {x + 2} \right)}}} \right]dx} \\
= \frac{1}{2}\int\limits_3^4 {\left[ {\frac{1}{{{x^2}}} - \frac{1}{2}.\left( {\frac{1}{x} - \frac{1}{{x + 2}}} \right)} \right]} dx\\
= \mathop {\left. {\left[ {\frac{{ - 1}}{{2x}} - \frac{1}{4}\ln x + \frac{1}{4}\ln \left( {x + 2} \right)} \right]} \right|}\nolimits_3^4 \\
= \frac{{ - 1}}{8} + \frac{1}{6} - \frac{1}{4}\ln 4 + \frac{1}{4}\ln 3 + \frac{1}{4}\ln 6 - \frac{1}{4}\ln 5\\
= \frac{1}{4}\ln \frac{9}{{10}} + \frac{1}{{24}}\\
\Rightarrow a + b - c = - 5
\end{array}\)
