Formelsammlung Mathematik: Unbestimmte Integrale rationaler Funktionen

Zurück zu Formelsammlung Mathematik
Zurück zu Formelsammlung Mathematik: Integrale

---

Das Integral einer rationalen Funktion lässt sich immer in geschlossener Form angeben, wenn dies für die Nullstellen der Nennerfunktion der Fall ist. Das Standardverfahren hierfür ist die Partialbruchzerlegung, durch die sich das Problem auf die Integration einiger weniger Grundtypen rationaler Funktionen zurückführen lässt.

Die folgenden Tabellen enthalten außer diesen Grundtypen zahlreiche andere Typen von rationalen Funktionen, sodass die Partialbruchzerlegung in vielen Fällen überflüssig sein wird.

Integrale, die ax + b enthalten

[1]

• $\int (ax+b)^{n}\mathrm {d} x={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\mbox{( }}n\neq -1{\mbox{)}}$

Beweis

(1)  $\int {\left({ax+b}\right)}^{n}dx=\int {z^{n}dx}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

in (1) eingesetzt

(5)  $\int {\left({ax+b}\right)}^{n}dx={\frac {1}{a}}\int {z^{n}dz}={\frac {1}{a}}\cdot {\frac {z^{n+1}}{n+1}}+C$

(5a)  $\int {\left({ax+b}\right)}^{n}dx={\frac {\left({ax+b}\right)^{n+1}}{a\left({n+1}\right)}}+C$

 

[2]

• $\int {\frac {\mathrm {d} x}{ax+b}}={\frac {1}{a}}\ln \left|ax+b\right|+C$

Beweis

(1)  $\int {\frac {dx}{ax+b}}=\int {\frac {dx}{z}}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

in (1) eingesetzt

(5)  $\int {\frac {dx}{ax+b}}={\frac {1}{a}}\int {\frac {dz}{z}}$

mit

$\int {\frac {dz}{z}}=\ln z+C$

(6)  $\int {\frac {dx}{ax+b}}={\frac {1}{a}}\ln z+C$

(6a)  $\int {\frac {dx}{ax+b}}={\frac {1}{a}}\ln \left({ax+b}\right)+C$

 

[3]

• $\int {x\left({ax+b}\right)^{n}dx}={\frac {ax\left({n+1}\right)-b}{a^{2}\left({n+1}\right)\left({n+2}\right)}}\left({ax+b}\right)^{n+1}+C$

Beweis

(1)  $\int {x\left({ax+b}\right)^{n}dx}=\int {xz^{n}dx}$

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {1}{a^{2}}}\int {\left({z-b}\right)z^{n}dz}={\frac {1}{a^{2}}}\left({\int {z^{n+1}dz-b\int {z^{n}dz}}}\right)$

(7)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {1}{a^{2}}}\left({{\frac {z^{n+2}}{n+2}}-b{\frac {z^{n+1}}{n+1}}}\right)+C$

(7a)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {1}{a^{2}}}\left[{{\frac {\left({ax+b}\right)^{n+2}}{n+2}}-b{\frac {\left({ax+b}\right)^{n+1}}{n+1}}}\right]+C$

(7b)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {\left({ax+b}\right)^{n+1}}{a^{2}}}\left[{\frac {\left({ax+b}\right)\left({n+1}\right)-b\left({n+2}\right)}{\left({n+1}\right)\left({n+2}\right)}}\right]+C$

(7c)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {\left({ax+b}\right)^{n+1}\left({axn+ax+bn+b-bn-2b}\right)}{a^{2}\left({n+1}\right)\left({n+2}\right)}}+C$

(7d)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {\left({ax+b}\right)^{n+1}\left({axn+ax-b}\right)}{a^{2}\left({n+1}\right)\left({n+2}\right)}}+C$

(7e)  $\int {x\left({ax+b}\right)^{n}dx}={\frac {ax\left({n+1}\right)-b}{a^{2}\left({n+1}\right)\left({n+2}\right)}}\left({ax+b}\right)^{n+1}+C$

 

[4]

• Nützlich ist auch die Formel
  $\int x^{m}(ax+b)^{n}\mathrm {d} x={\frac {1}{a^{m+1}}}\int {\left(X-b\right)^{m}X^{n}\mathrm {d} X}+C{\text{ mit }}X=ax+b\qquad {\mbox{(}}n\not \in \{-1,-2,\ldots ,-m\}{\mbox{)}},$
  wenn $m$ eine natürliche Zahl mit $m<n$ ist. Dann kann $\left(X-b\right)^{m}$ leichter nach dem binomischen Lehrsatz entwickelt werden.

Oder

$\int {x^{m}\left({ax+b}\right)^{n}dx}={\frac {1}{a^{m}}}\int {\left({ax}\right)^{m}\left({ax+b}\right)^{n}dx}+C$

Beweis

(1)  $\int {x^{m}\left({ax+b}\right)^{n}dx}=\int {x^{m}z^{n}dx}$

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)  $\int {x^{m}\left({ax+b}\right)^{n}dx}=\int {{\frac {\left({z-b}\right)^{m}}{a^{m}}}z^{n}{\frac {dz}{a}}}+C$

mit

$X=\left({ax+b}\right)$

und

$dX=d\left({ax+b}\right)=dz$

(6a)  $\int {x^{m}\left({ax+b}\right)^{n}dx}={\frac {1}{a^{m+1}}}\int {\left({X-b}\right)^{m}X^{n}dX+C}$

anders aufgelöst und (2) und (4) eingesetzt

(7)  $\int {x^{m}\left({ax+b}\right)^{n}dx}=\int {{\frac {\left({z-b}\right)^{m}}{a^{m}}}z^{n}{\frac {dz}{a}}}={\frac {1}{a^{m+1}}}\int {\left({ax+b-b}\right)^{m}\left({ax+b}\right)^{n}adx}+C$

(7a)  $\int {x^{m}\left({ax+b}\right)^{n}dx}={\frac {1}{a^{m}}}\int {\left({ax}\right)^{m}\left({ax+b}\right)^{n}dx}+C$

 

[5]

• $\int {\frac {x\;\mathrm {d} x}{ax+b}}={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|+C$

Beweis

(1)  $\int {{x \over {ax+b}}dx}=\int {{x \over z}dx}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)  $\int {{x \over {ax+b}}dx}={1 \over {a^{2}}}\int {{{z-b} \over z}dz}={1 \over {a^{2}}}\left({\int {dz-b\int {{dz} \over z}}}\right)$

mit

$\int {dz}=z+C$

und

$\int {{dz} \over z}=\ln z+C$

in (6) eingesetzt

(6a)  $\int {{x \over {ax+b}}dx}={1 \over {a^{2}}}\left({z-b\cdot \ln z}\right)+C$

(2) eingesetzt

(6b)  $\int {{x \over {ax+b}}dx}={1 \over {a^{2}}}\left[{ax+b-b\cdot \ln \left({ax+b}\right)}\right]+C$

mit

${b \over {a^{2}}}=C$

(6c)  $\int {{x \over {ax+b}}dx}={x \over a}-{b \over {a^{2}}}\ln \left({ax+b}\right)+C$

 

[6]

$\int {\frac {x\;\mathrm {d} x}{(ax+b)^{2}}}={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|+C$

Beweis

(1)  $\int {{x \over {\left({ax+b}\right)^{2}}}dx}=\int {{x \over {z^{2}}}dx}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)  $\int {{x \over {\left({ax+b}\right)^{2}}}dx}={1 \over {a^{2}}}\int {{{z-b} \over {z^{2}}}dz}={1 \over {a^{2}}}\left({\int {{{dz} \over z}-b\int {{dz} \over {z^{2}}}}}\right)$

mit

$\int {{dz} \over z}=\ln z+C$

und

$\int {{dz} \over {z^{2}}}=-{1 \over z}+C$

in (6) eingesetzt

(6a)  $\int {{x \over {\left({ax+b}\right)^{2}}}dx}={1 \over {a^{2}}}\left({\ln z+{b \over z}}\right)+C={1 \over {a^{2}}}\left[{\ln \left({ax+b}\right)+{b \over {ax+b}}}\right]+C$

oder

(6b)  $\int {{x \over {\left({ax+b}\right)^{2}}}dx}={b \over {a^{2}\left({ax+b}\right)}}+{1 \over {a^{2}}}\ln \left({ax+b}\right)+C$

 

[7]

$\int {{x \over {\left({ax+b}\right)^{n}}}dx}={1 \over {a^{2}}}\left[{{1 \over {\left({2-n}\right)\left({ax+b}\right)^{n-2}}}-b{1 \over {\left({1-n}\right)\left({ax+b}\right)^{n-1}}}}\right]+C\qquad {\mbox{(}}n\not \in \{1,2\}{\mbox{)}}$

Beweis

(1)  $\int {{x \over {\left({ax+b}\right)^{n}}}dx}=\int {{x \over {z^{n}}}dx}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)   $\int {{x \over {\left({ax+b}\right)^{n}}}dx}={1 \over {a^{2}}}\int {{{z-b} \over {z^{n}}}dz}={1 \over {a^{2}}}\left({\int {{{dz} \over {z^{n-1}}}-b\int {{dz} \over {z^{n}}}}}\right)$

mit

$\int {{dz} \over {z^{n-1}}}={1 \over {\left({2-n}\right)z^{n-2}}}+C$

und

$\int {\frac {dz}{z^{n}}}={\frac {1}{\left({1-n}\right)z^{n-1}}}+C$

in (6) eingesetzt

(6a)  $\int {{\frac {x}{\left({ax+b}\right)^{n}}}dx}={\frac {1}{a^{2}}}\left[{{\frac {1}{\left({2-n}\right)z^{n-2}}}-b{\frac {1}{\left({1-n}\right)z^{n-1}}}}\right]+C$

(6b)  $\int {{\frac {x}{\left({ax+b}\right)^{n}}}dx}={\frac {1}{a^{2}}}\left[{{\frac {1}{\left({2-n}\right)\left({ax+b}\right)^{n-2}}}-b{\frac {1}{\left({1-n}\right)\left({ax+b}\right)^{n-1}}}}\right]+C$

 

,

[8]

$\int {{{x^{2}} \over {ax+b}}dx}={x \over {2a^{2}}}\left({ax-2b}\right)+{{b^{2}} \over {a^{3}}}\ln \left({ax+b}\right)+C$

Beweis

(1)  $\int {{{x^{2}} \over {ax+b}}dx}=\int {{{x^{2}} \over z}dx}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)  $\int {{{x^{2}} \over {ax+b}}dx}={1 \over {a^{3}}}\int {{{\left({z-b}\right)^{2}} \over z}dz}={1 \over {a^{3}}}\int {{{z^{2}-2zb+b^{2}} \over z}dz}={1 \over {a^{3}}}\left({\int {zdz}-2b\int {dz+b^{2}\int {{dz} \over z}}}\right)$

mit

$\int {zdz}={{z^{2}} \over 2}+C$

und

$\int {dz}=z+C$

und

$\int {{dz} \over z}=\ln z+C$

in (6) eingesetzt

(6a)  $\int {{{x^{2}} \over {ax+b}}dx}={1 \over {a^{3}}}\left[{{{z^{2}} \over 2}-2bz+b^{2}\ln z}\right]+C$

(2) eingesetzt

(6b)  $\int {{{x^{2}} \over {ax+b}}dx}={1 \over {a^{3}}}\left[{{{\left({ax+b}\right)^{2}} \over 2}-2b\left({ax+b}\right)+b^{2}\ln \left({ax+b}\right)}\right]+C$

oder

(6c)  $\int {{{x^{2}} \over {ax+b}}dx}={{a^{2}x^{2}+2abx+b^{2}} \over {2a^{3}}}-{{4abx+4b^{2}} \over {2a^{3}}}+{{b^{2}} \over {a^{3}}}\ln \left({ax+b}\right)+C$

(6d)  $\int {{{x^{2}} \over {ax+b}}dx}={{a^{2}x^{2}-2abx-3b^{2}} \over {2a^{3}}}+{{b^{2}} \over {a^{3}}}\ln \left({ax+b}\right)+C$

mit

$-{{3b^{2}} \over {2a^{3}}}=C$

(6e)  $\int {{{x^{2}} \over {ax+b}}dx}={x \over {2a^{2}}}\left({ax-2b}\right)+{{b^{2}} \over {a^{3}}}\ln \left({ax+b}\right)+C$

 

[9]

$\int {\frac {x^{2}\;\mathrm {d} x}{(ax+b)^{2}}}={\frac {1}{a^{3}}}\left(ax-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)+C$

Beweis

(1)  $\int {{{x^{2}} \over {\left({ax+b}\right)^{2}}}dx}=\int {{{x^{2}} \over {z^{2}}}dx}$

mit

(2)  $z=ax+b\,$

differenziert

(3)  ${\frac {dz}{dx}}=a$

(4)  $dx={\frac {dz}{a}}$

und (2) nach x aufgelöst

(5)  $x={\frac {z-b}{a}}$

in (1) eingesetzt

(6)  $\int {{{x^{2}} \over {\left({ax+b}\right)^{2}}}dx}={1 \over {a^{3}}}\int {{{\left({z-b}\right)^{2}} \over {z^{2}}}dz}={1 \over {a^{3}}}\int {{{z^{2}-2zb+b^{2}} \over {z^{2}}}dz}={1 \over {a^{3}}}\left({\int {dz}-2b\int {{{dz} \over z}+b^{2}\int {{dz} \over {z^{2}}}}}\right)$

mit

$\int {dz}=z+C$

und

$\int {{dz} \over z}=\ln z+C$

und

$\int {{dz} \over {z^{2}}}=-{1 \over z}+C$

in (6) eingesetzt

(6a)  $\int {{\frac {x^{2}}{\left({ax+b}\right)^{2}}}dx}={\frac {1}{a^{3}}}\left({z-2b\cdot \ln z-{\frac {b^{2}}{z}}}\right)+C$

(2) eingesetzt

(6b)  $\int {{\frac {x^{2}}{\left({ax+b}\right)^{2}}}dx}={\frac {1}{a^{3}}}\left[{ax+b-2b\cdot \ln \left({ax+b}\right)-{\frac {b^{2}}{ax+b}}}\right]+C$

mit

${\frac {b}{a^{3}}}=C$

(6c)  $\int {{\frac {x^{2}}{\left({ax+b}\right)^{2}}}dx}={\frac {1}{a^{3}}}\left[{ax-2b\cdot \ln \left({ax+b}\right)-{\frac {b^{2}}{ax+b}}}\right]+C$

 

[10]

$\int {\frac {x^{2}\;\mathrm {d} x}{(ax+b)^{3}}}={\frac {1}{a^{3}}}\left(\ln \left|ax+b\right|+{\frac {2b}{ax+b}}-{\frac {b^{2}}{2(ax+b)^{2}}}\right)+C$

[11]

$\int {\frac {x^{2}\;\mathrm {d} x}{(ax+b)^{n}}}={\frac {1}{a^{3}}}\left(-{\frac {1}{(n-3)(ax+b)^{n-3}}}+{\frac {2b}{(n-2)(ax+b)^{n-2}}}-{\frac {b^{2}}{(n-1)(ax+b)^{n-1}}}\right)+C\qquad {\mbox{(}}n\not \in \{1,2,3\}{\mbox{)}}$

[12]

• $\int {\frac {x^{3}\;\mathrm {d} x}{ax+b}}={\frac {1}{a^{4}}}\left({\frac {(ax+b)^{3}}{3}}-{\frac {3b(ax+b)^{2}}{2}}+3b^{2}(ax+b)-b^{3}\ln {\left|ax+b\right|}\right)+C$

[13]

$\int {\frac {x^{3}\;\mathrm {d} x}{(ax+b)^{2}}}={\frac {1}{a^{4}}}\left({\frac {(ax+b)^{2}}{2}}-3b(ax+b)+3b^{2}\ln {\left|ax+b\right|}+{\frac {b^{3}}{ax+b}}\right)+C$

[14]

$\int {\frac {x^{3}\;\mathrm {d} x}{(ax+b)^{3}}}={\frac {1}{a^{4}}}\left(ax+b-3b\ln \left|ax+b\right|-{\frac {3b^{2}}{ax+b}}+{\frac {b^{3}}{2(ax+b)^{2}}}\right)+C$

[15]

$\int {\frac {x^{3}\;\mathrm {d} x}{(ax+b)^{4}}}={\frac {1}{a^{4}}}\left(\ln \left|ax+b\right|+{\frac {3b}{ax+b}}-{\frac {3b^{2}}{(ax+b)^{2}}}+{\frac {b^{3}}{3(ax+b)^{3}}}\right)+C$

[16]

${\begin{aligned}\int {\frac {x^{3}\;\mathrm {d} x}{(ax+b)^{n}}}=&{\frac {1}{a^{4}}}\left({\frac {-1}{(n-4)(ax+b)^{n-4}}}+{\frac {3b}{(n-3)(ax+b)^{n-3}}}-{\frac {3b^{2}}{(n-2)(ax+b)^{n-2}}}+{\frac {b^{3}}{(n-1)(ax+b)^{n-1}}}\right)+C\\&{\mbox{(}}n\not \in \{1,2,3,4\}{\mbox{)}}\end{aligned}}$

[17]

• $\int {\frac {\mathrm {d} x}{x(ax+b)}}=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|+C$

[18]

$\int {\frac {\mathrm {d} x}{x(ax+b)^{n}}}=-{\frac {1}{b^{n}}}\left[\ln \left|{\frac {ax+b}{x}}\right|-\sum _{i=1}^{n-1}{\binom {n-1}{i}}{\frac {(-a)^{i}x^{i}}{i(ax+b)^{i}}}\right]+C\qquad {\mbox{(}}n\geq 1{\mbox{)}}$
Für $n=1$ ist die Summe leer und es es entsteht die vorhergehende Formel.

[19]

• $\int {\frac {\mathrm {d} x}{x^{2}(ax+b)}}=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|+C$

[20]

$\int {\frac {\mathrm {d} x}{x^{2}(ax+b)^{2}}}=-a\left({\frac {1}{b^{2}(ax+b)}}+{\frac {1}{ab^{2}x}}-{\frac {2}{b^{3}}}\ln \left|{\frac {ax+b}{x}}\right|\right)+C$

[21]

$\int {\frac {\mathrm {d} x}{x^{2}(ax+b)^{3}}}=-a\left({\frac {1}{2b^{2}(ax+b)^{2}}}+{\frac {2}{b^{3}(ax+b)}}+{\frac {1}{ab^{3}x}}-{\frac {3}{b^{4}}}\ln \left|{\frac {ax+b}{x}}\right|\right)+C$

[22]

$\int {\frac {\mathrm {d} x}{x^{2}(ax+b)^{n}}}=-{\frac {1}{b^{n+1}}}\left[{\frac {ax+b}{x}}-na\ln \left|{\frac {ax+b}{x}}\right|-\sum _{i=2}^{n}{\binom {n}{i}}{\frac {(-a)^{i}x^{i-1}}{(i-1)(ax+b)^{i-1}}}\right]+C\qquad {\mbox{(}}n\geq 1{\mbox{)}}$
Für n = 1 (mit leerer Summe) und für n = 3 sind dies andere Stammfunktionen (beziehungsweise eine andere Konstante) als oben angegeben, für n = 2 ist es eine andere Darstellung der selben Stammfunktion.

[23]

• $\int {\frac {\mathrm {d} x}{x^{3}(ax+b)}}=-{\frac {1}{b^{3}}}\left[a^{2}\ln \left|{\frac {ax+b}{x}}\right|-{\frac {2a(ax+b)}{x}}+{\frac {(ax+b)^{2}}{2x^{2}}}\right]+C$

[24]

$\int {\frac {\mathrm {d} x}{x^{3}(ax+b)^{2}}}=-{\frac {1}{b^{4}}}\left[3a^{2}\ln \left|{\frac {ax+b}{x}}\right|+{\frac {a^{3}x}{ax+b}}+{\frac {(ax+b)^{2}}{2x^{2}}}-{\frac {3a(ax+b)}{x}}\right]+C$

[25]

$\int {\frac {\mathrm {d} x}{x^{3}(ax+b)^{3}}}=-{\frac {1}{b^{5}}}\left[6a^{2}\ln \left|{\frac {ax+b}{x}}\right|+{\frac {4a^{3}x}{ax+b}}-{\frac {a^{4}x^{2}}{2(ax+b)^{2}}}+{\frac {(ax+b)^{2}}{2x^{2}}}-{\frac {4a(ax+b)}{x}}\right]+C$

[26]

${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{3}(ax+b)^{n}}}=-{\frac {1}{b^{n+2}}}&\left[{\frac {n(n+1)}{2}}a^{2}\ln \left|{\frac {ax+b}{x}}\right|+{\frac {(ax+b)^{2}}{2x^{2}}}-{\frac {(n+1)a(ax+b)}{x}}\right.\\&\left.-\sum _{i=3}^{n+1}{\binom {n+1}{i}}{\frac {(-a)^{i}x^{i-2}}{(i-2)(ax+b)^{i-2}}}\right]+C\qquad {\mbox{(}}n\geq 1{\mbox{)}}\end{aligned}}$
Für n = 1 (wo die Summe leer ist), n = 2 und n = 3 sind dies die oben angegebenen Funktionen.

[27]

• ${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{m}(ax+b)^{n}}}=&-{\frac {1}{b^{m+n-1}}}\left[{\binom {m+n-2}{m-1}}(-a)^{m-1}\ln \left|{\frac {ax+b}{x}}\right|+\sum _{\begin{smallmatrix}{i=0}\\{i\neq m-1}\end{smallmatrix}}^{m+n-2}{\binom {m+n-2}{i}}{\frac {(-a)^{i}(ax+b)^{m-i-1}}{(m-i-1)x^{m-i-1}}}\right]+C\\&{\mbox{(}}m\geq 1,\,n\geq 1{\mbox{)}}\end{aligned}}$
  Diese Verallgemeinerung umfasst die meisten der oben angegebenen Formeln.

Integrale, die zwei Linearfaktoren enthalten

• $\int {\frac {ax+b}{cx+d}}\,\mathrm {d} x={\frac {ax}{c}}+{\frac {bc-ad}{c^{2}}}\ln \left|cx+d\right|+C$

• $\int {\frac {\mathrm {d} x}{(ax+b)(cx+d)}}={\frac {-1}{bc-ad}}\ln \left|{\frac {ax+b}{cx+d}}\right|+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

$\int {\frac {x\,\mathrm {d} x}{(ax+b)(cx+d)}}={\frac {1}{bc-ad}}\left({\frac {b}{a}}\ln \left|ax+b\right|-{\frac {d}{c}}\ln \left|cx+d\right|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

• $\int {\frac {\mathrm {d} x}{(ax+b)^{2}(cx+d)}}={\frac {1}{bc-ad}}\left({\frac {1}{ax+b}}-{\frac {c}{bc-ad}}\ln \left|{\frac {ax+b}{cx+d}}\right|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

$\int {\frac {x\,\mathrm {d} x}{(ax+b)^{2}(cx+d)}}={\frac {-b}{a(bc-ad)(ax+b)}}+{\frac {d}{(bc-ad)^{2}}}\left(\ln \left|{\frac {c}{a}}\right|+\ln \left|{\frac {ax+b}{cx+d}}\right|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

$\int {\frac {x^{2}\mathrm {d} x}{(ax+b)^{2}(cx+d)}}={\frac {1}{(bc-ad)^{2}}}\left({\frac {b^{2}(bc-ad)}{a^{2}(ax+b)}}+{\frac {b}{a^{2}}}(bc-2ad)\ln |ax+b|+{\frac {d^{2}}{c}}\ln |cx+d|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

• $\int {\frac {\mathrm {d} x}{(ax+b)^{2}(cx+d)^{2}}}=-{\frac {bc+ad+2acx}{(bc-ad)^{2}(ax+b)(cx+d)}}+{\frac {2ac}{(bc-ad)^{3}}}\left(\ln \left|{\frac {c}{a}}\right|+\ln \left|{\frac {ax+b}{cx+d}}\right|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

$\int {\frac {x\,\mathrm {d} x}{(ax+b)^{2}(cx+d)^{2}}}={\frac {2bd+(bc+ad)x}{(bc-ad)^{2}(ax+b)(cx+d)}}-{\frac {bc+ad}{(bc-ad)^{3}}}\left(\ln \left|{\frac {c}{a}}\right|+\ln \left|{\frac {ax+b}{cx+d}}\right|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

$\int {\frac {x^{2}\mathrm {d} x}{(ax+b)^{2}(cx+d)^{2}}}=-{\frac {bd(bc+ad)+(b^{2}c^{2}+a^{2}d^{2})x}{ac(bc-ad)^{2}(ax+b)(cx+d)}}+{\frac {2bd}{(bc-ad)^{3}}}\left(\ln \left|{\frac {c}{a}}\right|+\ln \left|{\frac {ax+b}{cx+d}}\right|\right)+C\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}$

Integrale, die x² ± a² enthalten

• $\int {\frac {\mathrm {d} x}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C$

Beweis

(1)  $\int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a^{2}}}\int {\frac {dx}{{\frac {x^{2}}{a^{2}}}+1}}={\frac {1}{a^{2}}}\int {\frac {dx}{z^{2}+1}}$

mit

(2)  $z={\frac {x}{a}}$

differenziert

(3)  ${\frac {dz}{dx}}={\frac {1}{a}}$

(4)  $dx=a\cdot dz\,$

und (2) nach x aufgelöst

(5)  $x=az\,$

in (1) eingesetzt

(6)  $\int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a^{2}}}\int {\frac {a\cdot dz}{z^{2}+1}}={\frac {1}{a}}\int {\frac {dz}{z^{2}+1}}$

mit Grundintegral

$\int {\frac {dz}{z^{2}+1}}=\arctan z+C$

in (6) eingesetzt

(7)  $\int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan z+C$

(2) eingesetzt

(8)  $\int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C$

 

${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{2}-a^{2}}}&={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C\\&={\begin{cases}-{\frac {1}{a}}\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\-{\frac {1}{a}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

Beweis

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

durch Partialbruchzerlegung

(2)  ${\frac {1}{\left({x-a}\right)\left({x+a}\right)}}={\frac {A}{x-a}}+{\frac {B}{x+a}}={\frac {A\left({x+a}\right)+B\left({x-a}\right)}{\left({x-a}\right)\left({x+a}\right)}}$

(2a)  $1=A\left({x+a}\right)+B\left({x-a}\right)$

gesetzt

(3a)  $x+a=0\,$

(3b)  $x=-a\,$

in (2a) eingesetzt

(3c)  $1=A\left({-a+a}\right)+B\left({-a-a}\right)=-2aB$

(3d)  $B=-{\frac {1}{2a}}$

gesetzt

(4a)  $x-a=0\,$

(4b)  $x=a\,$

in (2a) eingesetzt

(4c)  $1=A\left({a+a}\right)+B\left({a-a}\right)=2aA$

(4d)  $A={\frac {1}{2a}}$

(3d) und (4d) in (2) eingesetzt

(5)  ${\frac {1}{\left({x-a}\right)\left({x+a}\right)}}={\frac {1}{2a\left({x-a}\right)}}-{\frac {1}{2a\left({x+a}\right)}}$

Probe

${\frac {1}{2a\left({x-a}\right)}}-{\frac {1}{2a\left({x+a}\right)}}={\frac {x+a-x+a}{2a\left({x-a}\right)\left({x+a}\right)}}={\frac {1}{\left({x-a}\right)\left({x+a}\right)}}$

(5) in (1) eingesetzt

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

mit (siehe [1.2])

$\int {\frac {dx}{x-a}}=\ln \left({x-a}\right)+C$

und

$\int {\frac {dx}{x+a}}=\ln \left({x+a}\right)+C$

in (6) eingesetzt

(7)  $\int {\frac {dx}{x^{2}-a^{2}}}={\frac {1}{2a}}\left[{\ln \left({x-a}\right)-\ln \left({x+a}\right)}\right]+C$

nach Logarithmenregel

$\log u-\log v=\log {\frac {u}{v}}$

(8)  $\int {\frac {dx}{x^{2}-a^{2}}}={\frac {1}{2a}}\ln {\frac {x-a}{x+a}}+C$

 

$\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})^{2}}}={\frac {x}{2a^{2}(x^{2}+a^{2})}}+{\frac {1}{2a^{3}}}\arctan {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})^{2}}}&=-{\frac {x}{2a^{2}(x^{2}-a^{2})}}+{\frac {1}{4a^{3}}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&=-{\frac {x}{2a^{2}(x^{2}-a^{2})}}+{\begin{cases}{\frac {1}{2a^{3}}}\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{2a^{3}}}\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

$\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})^{3}}}={\frac {x}{4a^{2}(x^{2}+a^{2})^{2}}}+{\frac {3x}{8a^{4}(x^{2}+a^{2})}}+{\frac {3}{8a^{5}}}\arctan {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})^{3}}}&=-{\frac {x}{4a^{2}(x^{2}-a^{2})^{2}}}+{\frac {3x}{8a^{4}(x^{2}-a^{2})}}-{\frac {3}{16a^{5}}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&=-{\frac {x}{4a^{2}(x^{2}-a^{2})^{2}}}+{\frac {3x}{8a^{4}(x^{2}-a^{2})}}-{\begin{cases}{\frac {3}{8a^{5}}}\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {3}{8a^{5}}}\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

Rekursionsformeln:
$\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})^{n}}}={\frac {x}{2(n-1)a^{2}(x^{2}+a^{2})^{n-1}}}+{\frac {2n-3}{2(n-1)a^{2}}}\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})^{n-1}}}\qquad {\mbox{(}}n\geq 2{\mbox{)}}$

$\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})^{n}}}={\frac {-x}{2(n-1)a^{2}(x^{2}-a^{2})^{n-1}}}-{\frac {2n-3}{2(n-1)a^{2}}}\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})^{n-1}}}\qquad {\mbox{(}}n\geq 2{\mbox{)}}$

• $\int {\frac {x\;\mathrm {d} x}{x^{2}\pm a^{2}}}={\frac {1}{2}}\ln \left|x^{2}\pm a^{2}\right|+C$

$\int {\frac {x\;\mathrm {d} x}{(x^{2}\pm a^{2})^{2}}}=-{\frac {1}{2(x^{2}\pm a^{2})}}+C$

$\int {\frac {x\;\mathrm {d} x}{(x^{2}\pm a^{2})^{3}}}=-{\frac {1}{4(x^{2}\pm a^{2})^{2}}}+C$

$\int {\frac {x\;\mathrm {d} x}{(x^{2}\pm a^{2})^{n}}}=-{\frac {1}{2(n-1)(x^{2}\pm a^{2})^{n-1}}}+C\qquad {\mbox{(}}n\neq 1{\mbox{)}}$

• $\int {\frac {x^{2}\mathrm {d} x}{x^{2}+a^{2}}}=x-a\,\mathrm {arctan} {\frac {x}{a}}+C$

Beweis

durch Umformung

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

mit Grundintegral

$\int {dx=x+C}$

und Integral [3.1]

$\int {\frac {dx}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C$

in (1) eingesetzt

(2)  $\int {\frac {x^{2}}{x^{2}+a^{2}}}dx=x+a\arctan {\frac {x}{a}}+C$

 

${\begin{aligned}\int {\frac {x^{2}\mathrm {d} x}{x^{2}-a^{2}}}&=x-{\frac {a}{2}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&=x-{\begin{cases}a\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\a\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

$\int {\frac {x^{2}\mathrm {d} x}{(x^{2}+a^{2})^{2}}}={\frac {-x}{2(x^{2}+a^{2})}}+{\frac {1}{2a}}\,\mathrm {arctan} {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {x^{2}\mathrm {d} x}{(x^{2}-a^{2})^{2}}}&={\frac {-x}{2(x^{2}-a^{2})}}-{\frac {1}{4a}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&={\frac {-x}{2(x^{2}-a^{2})}}-{\begin{cases}{\frac {1}{2a}}\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{2a}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

$\int {\frac {x^{2}\mathrm {d} x}{(x^{2}+a^{2})^{3}}}={\frac {-x}{4(x^{2}+a^{2})^{2}}}+{\frac {x}{8a^{2}(x^{2}+a^{2})}}+{\frac {1}{8a^{3}}}\,\mathrm {arctan} {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {x^{2}\mathrm {d} x}{(x^{2}-a^{2})^{3}}}&={\frac {-x}{4(x^{2}-a^{2})^{2}}}-{\frac {x}{8a^{2}(x^{2}-a^{2})}}+{\frac {1}{16a^{3}}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&={\frac {-x}{4(x^{2}-a^{2})^{2}}}-{\frac {x}{8a^{2}(x^{2}-a^{2})}}+{\begin{cases}{\frac {1}{8a^{3}}}\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{8a^{3}}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

Rekursionsformel:
$\int {\frac {x^{2}\mathrm {d} x}{(x^{2}\pm a^{2})^{n}}}={\frac {-x}{2(n-1)(x^{2}\pm a^{2})^{n-1}}}+{\frac {1}{2(n-1)}}\int {\frac {\mathrm {d} x}{(x^{2}\pm a^{2})^{n-1}}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}$

• $\int {\frac {x^{3}\mathrm {d} x}{x^{2}\pm a^{2}}}={\frac {x^{2}}{2}}\mp {\frac {a^{2}}{2}}\ln \left|x^{2}\pm a^{2}\right|+C$

$\int {\frac {x^{3}\mathrm {d} x}{(x^{2}\pm a^{2})^{2}}}=\pm {\frac {a^{2}}{2(x^{2}\pm a^{2})}}+{\frac {1}{2}}\ln \left|x^{2}\pm a^{2}\right|+C$

$\int {\frac {x^{3}\mathrm {d} x}{(x^{2}\pm a^{2})^{3}}}={\frac {-1}{2(x^{2}\pm a^{2})}}\pm {\frac {a^{2}}{4(x^{2}\pm a^{2})^{2}}}+C$

$\int {\frac {x^{3}\mathrm {d} x}{(x^{2}\pm a^{2})^{n}}}={\frac {-1}{2(n-2)(x^{2}\pm a^{2})^{n-2}}}\pm {\frac {a^{2}}{2(n-1)(x^{2}\pm a^{2})^{n-1}}}+C\qquad {\mbox{(}}n>2{\mbox{)}}$

• $\int {\frac {\mathrm {d} x}{x(x^{2}\pm a^{2})}}=\pm {\frac {1}{2a^{2}}}\ln \left|{\frac {x^{2}}{x^{2}\pm a^{2}}}\right|+C$

$\int {\frac {\mathrm {d} x}{x(x^{2}\pm a^{2})^{2}}}=\pm {\frac {1}{2a^{2}(x^{2}\pm a^{2})}}+{\frac {1}{2a^{4}}}\ln \left|{\frac {x^{2}}{x^{2}\pm a^{2}}}\right|+C$

$\int {\frac {\mathrm {d} x}{x(x^{2}\pm a^{2})^{3}}}=\pm {\frac {1}{4a^{2}(x^{2}\pm a^{2})^{2}}}+{\frac {1}{2a^{4}(x^{2}\pm a^{2})}}\pm {\frac {1}{2a^{6}}}\ln \left|{\frac {x^{2}}{x^{2}\pm a^{2}}}\right|+C$

• $\int {\frac {\mathrm {d} x}{x^{2}(x^{2}+a^{2})}}={\frac {-1}{a^{2}x}}-{\frac {1}{a^{3}}}\,\mathrm {arctan} {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{2}(x^{2}-a^{2})}}&={\frac {1}{a^{2}x}}-{\frac {1}{2a^{3}}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&={\frac {1}{a^{2}x}}-{\begin{cases}{\frac {1}{a^{3}}}\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{a^{3}}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

$\int {\frac {\mathrm {d} x}{x^{2}(x^{2}+a^{2})^{2}}}={\frac {-1}{a^{4}x}}-{\frac {x}{2a^{4}(x^{2}+a^{2})}}-{\frac {3}{2a^{5}}}\,\mathrm {arctan} {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{2}(x^{2}-a^{2})^{2}}}&={\frac {-1}{a^{4}x}}-{\frac {x}{2a^{4}(x^{2}-a^{2})}}+{\frac {3}{4a^{5}}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&={\frac {-1}{a^{4}x}}-{\frac {x}{2a^{4}(x^{2}-a^{2})}}+{\begin{cases}{\frac {3}{2a^{5}}}\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {3}{2a^{5}}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

$\int {\frac {\mathrm {d} x}{x^{2}(x^{2}+a^{2})^{3}}}={\frac {-1}{a^{6}x}}-{\frac {x}{4a^{4}(x^{2}+a^{2})^{2}}}-{\frac {7x}{8a^{6}(x^{2}+a^{2})}}-{\frac {15}{8a^{7}}}\,\mathrm {arctan} {\frac {x}{a}}+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{2}(x^{2}-a^{2})^{3}}}&={\frac {1}{a^{6}x}}-{\frac {x}{4a^{4}(x^{2}-a^{2})^{2}}}+{\frac {7x}{8a^{6}(x^{2}-a^{2})}}-{\frac {15}{16a^{7}}}\ln \left|{\frac {x+a}{x-a}}\right|+C\\&={\frac {1}{a^{6}x}}-{\frac {x}{4a^{4}(x^{2}-a^{2})^{2}}}+{\frac {7x}{8a^{6}(x^{2}-a^{2})}}-{\begin{cases}{\frac {15}{8a^{7}}}\,\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {15}{8a^{7}}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

• $\int {\frac {\mathrm {d} x}{x^{3}(x^{2}\pm a^{2})}}=\mp {\frac {1}{2a^{2}x^{2}}}-{\frac {1}{2a^{4}}}\ln \left|{\frac {x^{2}}{x^{2}\pm a^{2}}}\right|+C$

$\int {\frac {\mathrm {d} x}{x^{3}(x^{2}\pm a^{2})^{2}}}={\frac {-1}{2a^{4}x^{2}}}-{\frac {1}{2a^{4}(x^{2}\pm a^{2})}}\mp {\frac {1}{a^{6}}}\ln \left|{\frac {x^{2}}{x^{2}\pm a^{2}}}\right|+C$

$\int {\frac {\mathrm {d} x}{x^{3}(x^{2}\pm a^{2})^{3}}}=\mp {\frac {1}{2a^{6}x^{2}}}\mp {\frac {1}{a^{6}(x^{2}\pm a^{2})}}-{\frac {1}{4a^{4}(x^{2}\pm a^{2})^{2}}}-{\frac {3}{2a^{8}}}\ln \left|{\frac {x^{2}}{x^{2}\pm a^{2}}}\right|+C$

Integrale, die x² ± a² und einen Linearfaktor enthalten

$\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})(bx+c)}}={\frac {1}{a^{2}b^{2}+c^{2}}}\left({\frac {b}{2}}\ln {\frac {(bx+c)^{2}}{x^{2}+a^{2}}}+{\frac {c}{a}}\arctan {\frac {x}{a}}\right)+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})(bx+c)}}&={\frac {1}{c^{2}-a^{2}b^{2}}}\left({\frac {b}{2}}\ln \left|{\frac {(bx+c)^{2}}{x^{2}-a^{2}}}\right|-{\frac {c}{2a}}\ln \left|{\frac {x+a}{x-a}}\right|\right)+C\\&={\frac {b}{2(c^{2}-a^{2}b^{2})}}\ln \left|{\frac {(bx+c)^{2}}{x^{2}-a^{2}}}\right|-{\begin{cases}{\frac {c}{a(c^{2}-a^{2}b^{2})}}\,\mathrm {artanh} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|<|a|\\{\frac {c}{a(c^{2}-a^{2}b^{2})}}\,\mathrm {arcoth} \,{\frac {x}{a}}+C\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}$

Integrale, die ax² + bx + c enthalten

• Grundlage bildet das Integral
  $\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}={\begin{cases}{\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C\qquad {\mbox{(}}4ac-b^{2}>0{\mbox{)}}\\{\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C\qquad {\mbox{(}}4ac-b^{2}<0{\mbox{)}}\end{cases}}$

Im Falle $4ac-b^{2}<0$ kann man auch dafür auch schreiben:
$\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}=-{\frac {2}{\sqrt {b^{2}-4ac}}}\ \cdot {\begin{cases}\mathrm {artanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C\qquad {\text{wenn }}\left|{\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\right|<1\\\mathrm {arcoth} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C\qquad {\text{wenn }}\left|{\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\right|>1\end{cases}}$

Dieses Integral wird im Rest dieses Abschnitts immer wieder verwendet, entweder direkt oder als Verankerung für eine Rekursion.

• $\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{2}}}={\frac {2ax+b}{(4ac-b^{2})(ax^{2}+bx+c)}}+{\frac {2a}{4ac-b^{2}}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

$\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{3}}}={\frac {2ax+b}{4ac-b^{2}}}\left({\frac {1}{2(ax^{2}+bx+c)^{2}}}+{\frac {3a}{(4ac-b^{2})(ax^{2}+bx+c)}}\right)+{\frac {6a^{2}}{(4ac-b^{2})^{2}}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

Rekursionsformel:
$\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}={\frac {2ax+b}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}+{\frac {2(2n-3)a}{(n-1)(4ac-b^{2})}}\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{n-1}}}+C\qquad {\mbox{(}}n>1{\mbox{)}}$
Für n = 2 und n = 3 sind das die oben angegebenen Stammfunktionen.

• $\int {\frac {x\;\mathrm {d} x}{ax^{2}+bx+c}}={\frac {1}{2a}}\ln \left|ax^{2}+bx+c\right|-{\frac {b}{2a}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

In geschlossener Form ist
$\int {\frac {ex+f}{ax^{2}+bx+c}}\mathrm {d} x={\frac {e}{2a}}\ln \left|ax^{2}+bx+c\right|+{\begin{cases}{\frac {2af-be}{a{\sqrt {4ac-b^{2}}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C\qquad {\mbox{(}}4ac-b^{2}>0{\mbox{)}}\\{\frac {2af-be}{2a{\sqrt {b^{2}-4ac}}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C\qquad {\mbox{(}}4ac-b^{2}<0{\mbox{)}}\end{cases}}$
(oder Schreibweise mit artanh und arcoth analog zu oben)

$\int {\frac {x\;\mathrm {d} x}{(ax^{2}+bx+c)^{2}}}=-{\frac {bx+2c}{(4ac-b^{2})(ax^{2}+bx+c)}}-{\frac {b}{4ac-b^{2}}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

Rekursionsformel:
$\int {\frac {x\;\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}=-{\frac {bx+2c}{(n-1)(4ac-b^{2})(ax^{2}+bx+c)^{n-1}}}-{\frac {(2n-3)b}{(n-1)(4ac-b^{2})}}\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{n-1}}}+C\qquad {\mbox{(}}n>1{\mbox{)}}$
Für n = 2 ist das die vorige Stammfunktion.

• $\int {\frac {x^{2}\mathrm {d} x}{ax^{2}+bx+c}}={\frac {x}{a}}-{\frac {b}{2a^{2}}}\ln \left|ax^{2}+bx+c\right|+{\frac {b^{2}-2ac}{2a^{2}}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

$\int {\frac {x^{2}\mathrm {d} x}{(ax^{2}+bx+c)^{2}}}={\frac {(b^{2}-2ac)x+bc}{a(4ac-b^{2})(ax^{2}+bx+c)}}+{\frac {2c}{4ac-b^{2}}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

Die folgende Formel gilt für alle ganzen Zahlen n. Für n = 1 und n = 2 handelt es sich um die eben genannten Stammfunktionen:
$\int {\frac {x^{2}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}={\frac {-x}{(2n-3)a(ax^{2}+bx+c)^{n-1}}}+{\frac {c}{(2n-3)a}}\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}-{\frac {(n-2)b}{(2n-3)a}}\int {\frac {x\;\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}+C$

• Die folgende sehr allgemeine Formel gilt für alle ganzen Zahlen $m$ und für alle $n\neq {\frac {m+1}{2}}$, also für nahezu alle Funktionen in diesem Abschnitt und zudem für eine Reihe von irrationalen Funktionen. Als Rekursionsformel brauchbar ist sie natürlich nur für $m>1$. Für negative $m$ gibt es eine andere Rekursionsformel, siehe unten.
  ${\begin{aligned}\int {\frac {x^{m}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}=&{\frac {-x^{m-1}}{(2n-m-1)a(ax^{2}+bx+c)^{n-1}}}+{\frac {(m-1)c}{(2n-m-1)a}}\int {\frac {x^{m-2}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}\\&-{\frac {(n-m)b}{(2n-m-1)a}}\int {\frac {x^{m-1}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}+C\qquad {\mbox{(}}m\neq 2n-1{\mbox{)}}\end{aligned}}$

Für m = 2n–1 gilt statt dessen:
$\int {\frac {x^{2n-1}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}={\frac {1}{a}}\int {\frac {x^{2n-3}\mathrm {d} x}{(ax^{2}+bx+c)^{n-1}}}-{\frac {c}{a}}\int {\frac {x^{2n-3}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}-{\frac {b}{a}}\int {\frac {x^{2n-2}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}+C$

• $\int {\frac {\mathrm {d} x}{x(ax^{2}+bx+c)}}={\frac {1}{2c}}\ln \left|{\frac {x^{2}}{ax^{2}+bx+c}}\right|-{\frac {b}{2c}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

Rekursionsformel:
$\int {\frac {\mathrm {d} x}{x(ax^{2}+bx+c)^{n}}}={\frac {1}{2c(n-1)(ax^{2}+bx+c)^{n-1}}}-{\frac {b}{2c}}\int {\frac {\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}+{\frac {1}{c}}\int {\frac {\mathrm {d} x}{x(ax^{2}+bx+c)^{n-1}}}+C\qquad {\mbox{(}}n>1{\mbox{)}}$

• $\int {\frac {\mathrm {d} x}{x^{2}(ax^{2}+bx+c)}}={\frac {b}{2c^{2}}}\ln \left|{\frac {ax^{2}+bx+c}{x^{2}}}\right|-{\frac {1}{cx}}+\left({\frac {b^{2}}{2c^{2}}}-{\frac {a}{c}}\right)\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

• Und schließlich wieder als Rekursionsformel:
  ${\begin{aligned}\int {\frac {\mathrm {d} x}{x^{m}(ax^{2}+bx+c)^{n}}}=&{\frac {-1}{(m-1)cx^{m-1}(ax^{2}+bx+c)^{n-1}}}-{\frac {(2n+m-3)a}{(m-1)c}}\int {\frac {\mathrm {d} x}{x^{m-2}(ax^{2}+bx+c)^{n}}}\\&-{\frac {(n+m-2)b}{(m-1)c}}\int {\frac {\mathrm {d} x}{x^{m-1}(ax^{2}+bx+c)^{n}}}+C\qquad {\mbox{(}}m>1{\mbox{)}}\end{aligned}}$

Integrale, die ax² + bx + c und einen Linearfaktor enthalten

• $\int {\frac {\mathrm {d} x}{(dx+e)(ax^{2}+bx+c)}}={\frac {d}{2(cd^{2}-bde+ae^{2})}}\ln \left|{\frac {(dx+e)^{2}}{ax^{2}+bx+c}}\right|+{\frac {2ae-bd}{2(cd^{2}-bde+ae^{2})}}\int {\frac {\mathrm {d} x}{ax^{2}+bx+c}}+C$

Integrale, die drei oder mehr Linearfaktoren enthalten

Die Werte a, b, c,… müssen alle voneinander verschieden sein.

• $\int {\frac {\mathrm {d} x}{(x+a)(x+b)(x+c)}}={\frac {\ln |x+a|}{(b-a)(c-a)}}+{\frac {\ln |x+b|}{(a-b)(c-b)}}+{\frac {\ln |x+c|}{(a-c)(b-c)}}+C$

${\begin{aligned}\int {\frac {\mathrm {d} x}{(x+a)(x+b)(x+c)(x+d)}}=&{\frac {\ln |x+a|}{(b-a)(c-a)(d-a)}}+{\frac {\ln |x+b|}{(a-b)(c-b)(d-b)}}+{\frac {\ln |x+c|}{(a-c)(b-c)(d-c)}}\\&+{\frac {\ln |x+d|}{(a-d)(b-d)(c-d)}}+C\end{aligned}}$

…und so weiter. Allgemein geschrieben sieht das so aus:
$\int {\frac {\mathrm {d} x}{\prod _{i=1}^{n}(x+a_{i})}}=\sum _{i=1}^{n}{\frac {\ln |x+a_{i}|}{\prod _{j=1 \atop {j\neq i}}^{n}(a_{j}-a_{i})}}+C\qquad {\mbox{(}}\forall _{i,j}\,a_{i}\neq a_{j}{\mbox{)}}$

Integrale, die x³ ± a³ enthalten

• $\int {\frac {\mathrm {d} x}{x^{3}\pm a^{3}}}={\frac {1}{6a^{2}}}\ln {\frac {(x\pm a)^{2}}{x^{2}\mp ax+a^{2}}}\pm {\frac {1}{a^{2}{\sqrt {3}}}}\arctan {\frac {2x\mp a}{a{\sqrt {3}}}}+C$
  Dieses Integral – und das übernächste – werden im Rest des Abschnitts mehrfach zur Abkürzung verwendet.

$\int {\frac {\mathrm {d} x}{(x^{3}\pm a^{3})^{2}}}=\pm {\frac {x}{3a^{3}(x^{3}\pm a^{3})}}\pm {\frac {2}{3a^{3}}}\int {\frac {\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

• $\int {\frac {x\,\mathrm {d} x}{x^{3}\pm a^{3}}}=\pm {\frac {1}{6a}}\ln {\frac {x^{2}\mp ax+a^{2}}{(x\pm a)^{2}}}+{\frac {1}{a{\sqrt {3}}}}\arctan {\frac {2x\mp a}{a{\sqrt {3}}}}+C$

$\int {\frac {x\,\mathrm {d} x}{(x^{3}\pm a^{3})^{2}}}=\pm {\frac {x^{2}}{3a^{3}(x^{3}\pm a^{3})}}\pm {\frac {1}{3a^{3}}}\int {\frac {x\,\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

• $\int {\frac {x^{2}\mathrm {d} x}{x^{3}\pm a^{3}}}=\pm {\frac {1}{3}}\ln |x^{3}\pm a^{3}|+C$

$\int {\frac {x^{2}\mathrm {d} x}{(x^{3}\pm a^{3})^{2}}}={\frac {-1}{3(x^{3}\pm a^{3})}}+C$

• $\int {\frac {x^{3}\mathrm {d} x}{x^{3}\pm a^{3}}}=x\mp a^{3}\int {\frac {\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

$\int {\frac {x^{3}\mathrm {d} x}{(x^{3}\pm a^{3})^{2}}}={\frac {-x}{3(x^{3}\pm a^{3})}}+{\frac {1}{3}}\int {\frac {\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

• $\int {\frac {\mathrm {d} x}{x(x^{3}\pm a^{3})}}=\pm {\frac {1}{3a^{3}}}\ln \left|{\frac {x^{3}}{x^{3}\pm a^{3}}}\right|+C$

$\int {\frac {\mathrm {d} x}{x(x^{3}\pm a^{3})^{2}}}=\pm {\frac {1}{3a^{3}(x^{3}\pm a^{3})}}+{\frac {1}{3a^{6}}}\ln \left|{\frac {x^{3}}{x^{3}\pm a^{3}}}\right|+C$

• $\int {\frac {\mathrm {d} x}{x^{2}(x^{3}\pm a^{3})}}=\mp {\frac {1}{a^{3}x}}\mp {\frac {1}{a^{3}}}\int {\frac {x\,\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

$\int {\frac {\mathrm {d} x}{x^{2}(x^{3}\pm a^{3})^{2}}}={\frac {-1}{a^{6}x}}-{\frac {x^{2}}{3a^{6}(x^{3}\pm a^{3})}}-{\frac {4}{3a^{6}}}\int {\frac {x\,\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

• $\int {\frac {\mathrm {d} x}{x^{3}(x^{3}\pm a^{3})}}=\mp {\frac {1}{2a^{3}x^{2}}}\mp {\frac {1}{a^{3}}}\int {\frac {\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

$\int {\frac {\mathrm {d} x}{x^{3}(x^{3}\pm a^{3})^{2}}}={\frac {-1}{2a^{6}x^{2}}}-{\frac {x}{3a^{6}(x^{3}\pm a^{3})}}-{\frac {5}{3a^{6}}}\int {\frac {\mathrm {d} x}{x^{3}\pm a^{3}}}+C$

Integrale, die x⁴ ± a⁴ enthalten

• $\int {\frac {\mathrm {d} x}{x^{4}+a^{4}}}={\frac {1}{4{\sqrt {2}}\,a^{3}}}\left(\ln {\frac {x^{2}+{\sqrt {2}}\,ax+a^{2}}{x^{2}-{\sqrt {2}}\,ax+a^{2}}}+4\arctan {\frac {{\sqrt {2}}\,ax}{{\sqrt {a^{4}+x^{4}}}+a^{2}-x^{2}}}\right)+C$

$\int {\frac {\mathrm {d} x}{x^{4}-a^{4}}}={\frac {-1}{4a^{3}}}\left(\ln \left|{\frac {x+a}{x-a}}\right|+2\arctan {\frac {x}{a}}\right)+C$

• $\int <!--\; \int \; -->\frac{x\mathrm{d}x}{x^{}}$