Đáp án:
\[a + 2b = \frac{5}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
t = {x^2} \Rightarrow \left\{ \begin{array}{l}
dt = 2xdx\\
x = 1 \Rightarrow t = 1\\
x = 2 \Rightarrow 4
\end{array} \right.\\
\int\limits_1^2 {\frac{{dx}}{{{x^5} + {x^3}}}} \\
= \int\limits_1^2 {\frac{{xdx}}{{{x^6} + {x^4}}}} \\
= \int\limits_1^4 {\frac{{\frac{{dt}}{2}}}{{{t^3} + {t^2}}}} \\
= \frac{1}{2}\int\limits_1^4 {\frac{{dt}}{{{t^2}\left( {t + 1} \right)}}} \\
= \frac{1}{2}.\int\limits_1^4 {\frac{{\left( {t + 1} \right) - t}}{{{t^2}\left( {t + 1} \right)}}dt} \\
= \frac{1}{2}.\int\limits_1^4 {\left( {\frac{1}{{{t^2}}} - \frac{1}{{t\left( {t + 1} \right)}}} \right)dt} \\
= \frac{1}{2}\int\limits_1^4 {\left( {\frac{1}{{{t^2}}} - \frac{{\left( {t + 1} \right) - t}}{{t\left( {t + 1} \right)}}} \right)dt} \\
= \frac{1}{2}\int\limits_1^4 {\left[ {\frac{1}{{{t^2}}} - \frac{1}{t} + \frac{1}{{t + 1}}} \right]dt} \\
= \frac{1}{2}.\mathop {\left. {\left( {\frac{{ - 1}}{t} - \ln \left| t \right| + \ln \left| {t + 1} \right|} \right)} \right|}\nolimits_1^4 \\
= \frac{1}{2}.\left( {\frac{3}{4} - \ln 4 + \ln 1 + \ln 5 - \ln 2} \right)\\
= \frac{3}{8} + \frac{1}{2}.\left( {\ln 5 - \left( {\ln 4 + \ln 2} \right)} \right)\\
= \frac{3}{8} + \frac{1}{2}\ln \frac{5}{8}\\
\Rightarrow a + 2b = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}
\end{array}\)
