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Пр.4)   Решити  $\int {\frac{{dx}}{{{x^5}2{x^3} + x}}} $

Пр.4)   $\int {\frac{{dx}}{{{x^5}2{x^3} + x}}} $

$\frac{1}{{{x^5} + 2{x^3} + x}} = \frac{{2{x^2} - 5x + 1}}{{x\left( {{x^4} + 2{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}} = $

$ = \frac{{A{{\left( {{x^2} + 1} \right)}^2} + \left( {Bx + C} \right)x\left( {{x^2} + 1} \right) + \left( {Dx + E} \right)x}}{{x{{\left( {{x^2} + 1} \right)}^2}}}$

$1 = A{\left( {{x^2} + 1} \right)^2} + \left( {Bx + C} \right)x\left( {{x^2} + 1} \right) + \left( {Dx + E} \right)x$

$0{x^4} + 0{x^3} + 0{x^2} + 0x + 1 = A{\left( {{x^2} + 1} \right)^2} + \left( {Bx + C} \right)x\left( {{x^2} + 1} \right) + \left( {Dx + E} \right)x = $

$ = A{x^4} + 2A{x^2} + A + B{x^4} + B{x^2} + C{x^3} + Cx + D{x^2} + Ex = $

$ = \left( {A + B} \right){x^4} + C{x^3} + \left( {2A + B + D} \right){x^2} + \left( {C + E} \right)x + A$

\[\left\{ \begin{gathered}
A + B = 0 \hfill \\
C = 0 \hfill \\
2A + B + D = 0 \hfill \\
C + E = 0 \hfill \\
A = 1 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
B = - 1 \hfill \\
C = 0 \hfill \\
D = - 1 \hfill \\
E = 0 \hfill \\
A = 1 \hfill \\
\end{gathered} \right.\]

$\int {\frac{{dx}}{{{x^5} + 2{x^3} + x}}}  = \int {\left( {\frac{1}{x} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)} dx = $

$ = \int {\frac{1}{x}dx - \int {\frac{x}{{{x^2} + 1}}} dx - \int {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} } dx = $

${x^2} + 1 = t$

$2xdx = dt$

$2xdx = dt$

$ = \int {\frac{1}{x}dx - \int {\frac{{\frac{{dt}}{2}}}{t}} dt - \int {\frac{{\frac{{dt}}{2}}}{{{t^2}}}} } dt = \int {\frac{1}{x}dx - \frac{1}{2}\int {\frac{{dt}}{t}} dt - \frac{1}{2}\int {{t^{ - 2}}} } dt = $

$ = \ln \left| x \right| - \frac{1}{2}\ln \left| t \right| - \frac{1}{2}\frac{{{t^{ - 1}}}}{{ - 1}} + C = \ln \left| x \right| - \ln {\left| t \right|^{\frac{1}{2}}} + \frac{1}{{2t}} + C = $

$ = \ln \left| x \right| - \ln {\left| {{x^2} + 1} \right|^{\frac{1}{2}}} + \frac{1}{{2\left( {{x^2} + 1} \right)}} + C = \ln \left| {\frac{x}{{{{\left( {{x^2} + 1} \right)}^{\frac{1}{2}}}}}} \right| + \frac{1}{{2\left( {{x^2} + 1} \right)}} + C = $

$ = \ln \left| {\frac{x}{{^{\sqrt {{x^2} + 1} }}}} \right| + \frac{1}{{2\left( {{x^2} + 1} \right)}} + C$