\[\begin{array}{l}
I = \int_1^2 {\frac{{dx}}{{{x^5} + {x^3}}}} \\
\frac{1}{{{x^5} + {x^3}}} = \frac{1}{{{x^3}({x^2} + 1)}} = \frac{{Ax + B}}{{{x^2} + 1}} + \frac{C}{x} + \frac{D}{{{x^2}}} + \frac{E}{{{x^3}}}\\
\Leftrightarrow {x^3}(Ax + B) + C{x^2}({x^2} + 1) + Dx({x^2} + 1) + E({x^2} + 1) = 1\\
x = 0 \Rightarrow E = 1\\
x = 1 \Rightarrow A + B + 2C + 2D = - 1\\
x = - 1 \Rightarrow A - B + 2C - 2D = - 1\\
\left\{ \begin{array}{l}
A + B + 2C + 2D = - 1\\
A - B + 2C - 2D = - 1
\end{array} \right. \Rightarrow 2A + 4C = - 2\\
x = 2 \Rightarrow 16A + 8B + 20C + 10D = - 4\\
x = - 2 \Rightarrow 16A - 8B + 20C - 10D = - 4\\
\left\{ \begin{array}{l}
16A + 8B + 20C + 10D = - 4\\
16A - 8B + 20C - 10D = - 4
\end{array} \right. \Rightarrow 32A + 40C = - 8\\
\left\{ \begin{array}{l}
2A + 4C = - 2\\
32A + 40C = - 8
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 1\\
C = - 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
B + 2D = 0\\
- 8B - 10D = 0
\end{array} \right. \Rightarrow B = D = 0\\
\frac{1}{{{x^5} + {x^3}}} = \frac{x}{{{x^2} + 1}} - \frac{1}{x} + \frac{1}{{{x^3}}} = \frac{1}{2}.\frac{{2x}}{{{x^2} + 1}} - \frac{1}{x} + {x^{ - 3}}\\
I = \int_1^2 {\left( {\frac{1}{2}.\frac{{2x}}{{{x^2} + 1}} - \frac{1}{x} + {x^{ - 3}}} \right)dx} = \left. {\left( {\frac{1}{2}\ln \left| {{x^2} + 1} \right| - \ln \left| x \right| - \frac{1}{{2{x^2}}}} \right)} \right|_1^2\\
I = \frac{1}{2}\ln 5 - \ln 2 - \frac{1}{8} - \frac{1}{2}\ln 2 + \ln 1 + \frac{1}{2} = \frac{1}{2}\ln \frac{5}{2} - \ln 2 + \frac{3}{8}
\end{array}\]
\[\begin{array}{l}
I = \int_0^1 {\frac{{xdx}}{{{{(x + 1)}^3}}}} = \int_0^1 {\frac{{(x + 1 - 1)dx}}{{{{(x + 1)}^3}}}} = \int_0^1 {\frac{{dx}}{{{{(x + 1)}^2}}} - \int_0^1 {{{(x + 1)}^{ - 3}}dx} } \\
I = \left. {\left( { - \frac{1}{{x + 1}}} \right)} \right|_0^1 - \left. {\left( {\frac{1}{{2{{(x + 1)}^2}}}} \right)} \right|_0^1 = \left. {\left( { - \frac{1}{{x + 1}} - \frac{1}{{2{{(x + 1)}^2}}}} \right)} \right|_0^1\\
I = \left( { - \frac{1}{2} - \frac{1}{8}} \right) - \left( { - 1 - \frac{1}{2}} \right) = \frac{7}{8}
\end{array}\]
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