TU Wien:Mathematik 2 UE (diverse)/Übungen SS07/Beispiel 33

Man berechne $\int {\frac {x^{4}+x^{2}-1}{(x-1)^{2}(x^{2}+2x+3)}}\,dx$

Lösungsvorschlag von leiwand

Zunächst ist eine Partialbruchzerlegung durchzuführen. Da bei der Partialbruchzerlegung das Nennerpolynom nicht größer als das Zählerpolynom sein darf muss dieses zuerst reduziert werden:

Nun kann für den verbleibenden Bruch die Partialbruchzerlegung durchgeführt werden, wobei wegen der doppelten Nullstelle bei $(x-1)$ die Terme wie folgt lauten:

${\begin{aligned}{\frac {x^{2}+4x-4}{(x-1)^{2}(x^{2}+2x+3)}}&={\frac {A}{x-1}}+{\frac {B}{(x-1)^{2}}}+{\frac {Cx+D}{x^{2}+2x+3}}\\x^{2}+4x-4&=A(x-1)(x^{2}+2x+3)+B(x^{2}+2x+3)+(Cx+D)(x-1)^{2}\\&=(A+C)x^{3}+(A+B-2C+D)x^{2}+(A+2B+C-2D)x-3A+3B+D\end{aligned}}$

Der Koeffizentenvergleich ergibt das folgende Gleichungssystem:

für $x^{3}:0=A+C$

für $x^{2}:1=A+B-2C+D$

für $x^{1}:4=A+2B+C-2D$

für $x^{0}:-4=-3A+3B+D$

Berechnung:

$x^{3}:A=-C$

$x^{1}:B=2+D$

$x^{2}:3C=2D+1$

$x^{0}:-10=6D+1$

Ergebnis:

$A={\frac {8}{9}};B={\frac {1}{6}};C=-{\frac {8}{9}};D=-{\frac {11}{6}}$

Das Problem kann somit umgeschrieben werden zu:

${\begin{aligned}\int {\frac {x^{4}+x^{2}-1}{(x-1)^{2}(x^{2}+2x+3)}}\,dx&=\int {\frac {x^{2}+4x-4}{(x-1)^{2}(x^{2}+2x+3)}}+1\,dx\\&=\int {\frac {\frac {8}{9}}{x-1}}+{\frac {\frac {1}{6}}{(x-1)^{2}}}+{\frac {-{\frac {8}{9}}x-{\frac {11}{6}}}{x^{2}+2x+3}}+1\,dx\\&={\frac {8}{9}}\int {\frac {1}{x-1}}\,dx+{\frac {1}{6}}\int {\frac {1}{(x-1)^{2}}}\,dx-{\frac {8}{9}}\int {\frac {x}{x^{2}+2x+3}}\,dx-{\frac {11}{6}}\int {\frac {1}{x^{2}+2x+3}}\,dx+\int 1\,dx\end{aligned}}$

Betrachten wir die Integrale getrennt voneinander:

Wegen $\int {\frac {f'(x)}{f(x)}}\,dx=\ln |f(x)|$ erhalten wir

$\int {\frac {1}{x-1}}\,dx=\ln |x-1|$

${\begin{aligned}\int {\frac {1}{(x-1)^{2}}}\,dx&=\int {\frac {1}{u^{2}}}\,du&{\text{mit }}u=(x-1),dx=du\\&=-{\frac {1}{u}}\\&=-{\frac {1}{x-1}}\\&={\frac {1}{1-x}}\end{aligned}}$

${\begin{aligned}\int {\frac {x}{(x^{2}+2x+3)}}\,dx&=\int {\frac {x+1-1}{(x+1)^{2}+2}}\,dx\\&=\int {\frac {x+1}{(x+1)^{2}+2}}\,dx-\int {\frac {1}{(x+1)^{2}+2}}\,dx\end{aligned}}$

Das können wir wieder in zwei Teilen lösen:

${\begin{aligned}\int {\frac {x+1}{(x+1)^{2}+2}}\,dx&=\int {\frac {u}{u^{2}+2}}\,du&{\text{mit }}u=x+1,dx=du\\&={\frac {1}{2}}\int {\frac {1}{v}}\,dv&{\text{mit }}v=u^{2}+2,du={\frac {dv}{2u}}\\&={\frac {1}{2}}\ln |v|\\&={\frac {1}{2}}\ln |u^{2}+2|\\&={\frac {1}{2}}\ln |(x+1)^{2}+2|\end{aligned}}$

${\begin{aligned}\int {\frac {1}{(x^{2}+2x+3)}}\,dx&=\int {\frac {1}{(x+1)^{2}+2}}\,dx\\&={\frac {1}{2}}\int {\frac {1}{({\frac {x+1}{\sqrt {2}}})^{2}+1}}\,dx\\&={\frac {\sqrt {2}}{2}}\int {\frac {1}{u^{2}+1}}\,du&{\text{mit }}u={\frac {x+1}{\sqrt {2}}},dx={\sqrt {2}}du\\&={\frac {1}{\sqrt {2}}}\arctan u\\&={\frac {1}{\sqrt {2}}}\arctan {\frac {x+1}{\sqrt {2}}}\end{aligned}}$

Damit kämen wir zur Gesamtlösung:

${\begin{aligned}\int {\frac {x^{4}+x^{2}-1}{(x-1)^{2}(x^{2}+2x+3)}}\,dx&={\frac {8}{9}}\ln |x-1|+{\frac {1}{6}}{\frac {1}{1-x}}-{\frac {8}{9}}{\frac {1}{2}}\ln |(x+1)^{2}+2|+\left({\frac {8}{9}}-{\frac {11}{6}}\right){\frac {1}{\sqrt {2}}}\arctan {\frac {x+1}{\sqrt {2}}}+x+c\\&={\frac {8}{9}}\ln |x-1|+{\frac {1}{6-6x}}-{\frac {4}{9}}\ln |(x+1)^{2}+2|-{\frac {17}{18{\sqrt {2}}}}\arctan {\frac {x+1}{\sqrt {2}}}+x+c\end{aligned}}$

Links

Integrale Schritt für Schritt: https://web.archive.org/web/20180817162910/http://www.wolframalpha.com/