Inhaltsverzeichnis
1 Integrale rationaler Funktionen
1.1 Integrale, die ax + b enthalten 1.2 Integrale, die zwei Linearfaktoren enthalten 1.3 Integrale, die x2 ± a2 enthalten 1.4 Integrale, die x2 ± a2 und einen Linearfaktor enthalten 1.5 Integrale, die ax2 + bx + c enthalten 1.6 Integrale, die ax2 + bx + c und einen Linearfaktor enthalten 1.7 Integrale, die drei oder mehr Linearfaktoren enthalten 1.8 Integrale, die x3 ± a3 enthalten 1.9 Integrale, die x4 ± a4 enthalten 1.10 Integrale, die zwei quadratische Faktoren enthalten
Integrale rationaler Funktionen
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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.
Integrale, die ax + b enthalten
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{\displaystyle \int (ax+b)^{n}\mathrm {d} x={\frac {(ax+b)^{n+1}}{a(n+1)}}\qquad {\mbox{( }}n\neq -1{\mbox{)}}}
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{\displaystyle \int {\frac {\mathrm {d} x}{ax+b}}={\frac {1}{a}}\ln \left|ax+b\right|}
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{\displaystyle \int x(ax+b)^{n}\mathrm {d} x={\frac {a(n+1)x-b}{a^{2}(n+1)(n+2)}}(ax+b)^{n+1}\qquad {\mbox{(}}n\not \in \{-1,-2\}{\mbox{)}}}
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{\displaystyle \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}{\text{ mit }}X=ax+b\qquad {\mbox{(}}n\not \in \{-1,-2,\ldots ,-m\}{\mbox{)}},}
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{\displaystyle \int {\frac {x\;\mathrm {d} x}{ax+b}}={\frac {x}{a}}-{\frac {b}{a^{2}}}\ln \left|ax+b\right|}
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{\displaystyle \int {\frac {x\;\mathrm {d} x}{(ax+b)^{2}}}={\frac {b}{a^{2}(ax+b)}}+{\frac {1}{a^{2}}}\ln \left|ax+b\right|}
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{\displaystyle \int {\frac {x\;\mathrm {d} x}{(ax+b)^{n}}}={\frac {1}{a^{2}}}\left({\frac {-1}{(n-2)(ax+b)^{n-2}}}+{\frac {b}{(n-1)(ax+b)^{n-1}}}\right)\qquad {\mbox{(}}n\not \in \{-1,-2\}{\mbox{)}}}
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{\displaystyle \int {\frac {x^{2}\;\mathrm {d} x}{ax+b}}={\frac {1}{a^{3}}}\left({\frac {(ax+b)^{2}}{2}}-2b(ax+b)+b^{2}\ln \left|ax+b\right|\right)}
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{\displaystyle \int {\frac {x^{2}\;\mathrm {d} x}{(ax+b)^{2}}}={\frac {1}{a^{3}}}\left(ax+b-2b\ln \left|ax+b\right|-{\frac {b^{2}}{ax+b}}\right)}
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{\displaystyle \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)}
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{\displaystyle \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)\qquad {\mbox{(}}n\not \in \{1,2,3\}{\mbox{)}}}
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{\displaystyle \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)}
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{\displaystyle \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)}
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{\displaystyle \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)}
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{\displaystyle \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)}
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{\displaystyle {\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)\\&{\mbox{(}}n\not \in \{1,2,3,4\}{\mbox{)}}\end{aligned}}}
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{\displaystyle \int {\frac {\mathrm {d} x}{x(ax+b)}}=-{\frac {1}{b}}\ln \left|{\frac {ax+b}{x}}\right|}
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{\displaystyle \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]\qquad {\mbox{(}}n\geq 1{\mbox{)}}}
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{\displaystyle n=1}
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{\displaystyle \int {\frac {\mathrm {d} x}{x^{2}(ax+b)}}=-{\frac {1}{bx}}+{\frac {a}{b^{2}}}\ln \left|{\frac {ax+b}{x}}\right|}
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{\displaystyle \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)}
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{\displaystyle \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)}
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{\displaystyle \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]\qquad {\mbox{(}}n\geq 1{\mbox{)}}}
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{\displaystyle \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]}
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{\displaystyle \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]}
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{\displaystyle \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]}
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{\displaystyle {\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]\qquad {\mbox{(}}n\geq 1{\mbox{)}}\end{aligned}}}
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{\displaystyle {\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]\\&{\mbox{(}}m\geq 1,\,n\geq 1{\mbox{)}}\end{aligned}}}
Integrale, die zwei Linearfaktoren enthalten
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{\displaystyle \int {\frac {ax+b}{cx+d}}\,\mathrm {d} x={\frac {ax}{c}}+{\frac {bc-ad}{c^{2}}}\ln \left|cx+d\right|}
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{\displaystyle \int {\frac {\mathrm {d} x}{(ax+b)(cx+d)}}={\frac {-1}{bc-ad}}\ln \left|{\frac {ax+b}{cx+d}}\right|\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
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{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
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{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
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d
|
)
(
b
c
−
a
d
≠
0
)
{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
∫
x
2
d
x
(
a
x
+
b
)
2
(
c
x
+
d
)
=
1
(
b
c
−
a
d
)
2
(
b
2
(
b
c
−
a
d
)
a
2
(
a
x
+
b
)
+
b
a
2
(
b
c
−
2
a
d
)
ln
|
a
x
+
b
|
+
d
2
c
ln
|
c
x
+
d
|
)
(
b
c
−
a
d
≠
0
)
{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
∫
d
x
(
a
x
+
b
)
2
(
c
x
+
d
)
2
=
−
b
c
+
a
d
+
2
a
c
x
(
b
c
−
a
d
)
2
(
a
x
+
b
)
(
c
x
+
d
)
+
2
a
c
(
b
c
−
a
d
)
3
(
ln
|
c
a
|
+
ln
|
a
x
+
b
c
x
+
d
|
)
(
b
c
−
a
d
≠
0
)
{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
∫
x
d
x
(
a
x
+
b
)
2
(
c
x
+
d
)
2
=
2
b
d
+
(
b
c
+
a
d
)
x
(
b
c
−
a
d
)
2
(
a
x
+
b
)
(
c
x
+
d
)
−
b
c
+
a
d
(
b
c
−
a
d
)
3
(
ln
|
c
a
|
+
ln
|
a
x
+
b
c
x
+
d
|
)
(
b
c
−
a
d
≠
0
)
{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
∫
x
2
d
x
(
a
x
+
b
)
2
(
c
x
+
d
)
2
=
−
b
d
(
b
c
+
a
d
)
+
(
b
2
c
2
+
a
2
d
2
)
x
a
c
(
b
c
−
a
d
)
2
(
a
x
+
b
)
(
c
x
+
d
)
+
2
b
d
(
b
c
−
a
d
)
3
(
ln
|
c
a
|
+
ln
|
a
x
+
b
c
x
+
d
|
)
(
b
c
−
a
d
≠
0
)
{\displaystyle \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)\qquad {\mbox{(}}bc-ad\neq 0{\mbox{)}}}
Integrale, die x2 ± a2 enthalten
Bearbeiten
∫
d
x
x
2
+
a
2
=
1
a
arctan
x
a
{\displaystyle \int {\frac {\mathrm {d} x}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}}
∫
d
x
x
2
−
a
2
=
−
1
2
a
ln
|
x
+
a
x
−
a
|
=
{
−
1
a
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
−
1
a
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\begin{aligned}\int {\frac {\mathrm {d} x}{x^{2}-a^{2}}}&={\frac {-1}{2a}}\ln \left|{\frac {x+a}{x-a}}\right|\\&={\begin{cases}-{\frac {1}{a}}\,\mathrm {artanh} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|<|a|\\-{\frac {1}{a}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
d
x
(
x
2
+
a
2
)
2
=
x
2
a
2
(
x
2
+
a
2
)
+
1
2
a
3
arctan
x
a
+
C
{\displaystyle \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}
∫
d
x
(
x
2
−
a
2
)
2
=
−
x
2
a
2
(
x
2
−
a
2
)
+
1
4
a
3
ln
|
x
+
a
x
−
a
|
=
−
x
2
a
2
(
x
2
−
a
2
)
+
{
1
2
a
3
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
1
2
a
3
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&=-{\frac {x}{2a^{2}(x^{2}-a^{2})}}+{\begin{cases}{\frac {1}{2a^{3}}}\mathrm {artanh} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{2a^{3}}}\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
d
x
(
x
2
+
a
2
)
3
=
x
4
a
2
(
x
2
+
a
2
)
2
+
3
x
8
a
4
(
x
2
+
a
2
)
+
3
8
a
5
arctan
x
a
+
C
{\displaystyle \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}
∫
d
x
(
x
2
−
a
2
)
3
=
−
x
4
a
2
(
x
2
−
a
2
)
2
+
3
x
8
a
4
(
x
2
−
a
2
)
−
3
16
a
5
ln
|
x
+
a
x
−
a
|
=
−
x
4
a
2
(
x
2
−
a
2
)
2
+
3
x
8
a
4
(
x
2
−
a
2
)
−
{
3
8
a
5
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
3
8
a
5
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&=-{\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}}\qquad {\text{wenn }}|x|<|a|\\{\frac {3}{8a^{5}}}\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
Rekursionsformeln:
∫
d
x
(
x
2
+
a
2
)
n
=
x
2
(
n
−
1
)
a
2
(
x
2
+
a
2
)
n
−
1
+
2
n
−
3
2
(
n
−
1
)
a
2
∫
d
x
(
x
2
+
a
2
)
n
−
1
(
n
≥
2
)
{\displaystyle \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{)}}}
∫
d
x
(
x
2
−
a
2
)
n
=
−
x
2
(
n
−
1
)
a
2
(
x
2
−
a
2
)
n
−
1
−
2
n
−
3
2
(
n
−
1
)
a
2
∫
d
x
(
x
2
−
a
2
)
n
−
1
(
n
≥
2
)
{\displaystyle \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{)}}}
∫
x
d
x
x
2
±
a
2
=
1
2
ln
|
x
2
±
a
2
|
{\displaystyle \int {\frac {x\;\mathrm {d} x}{x^{2}\pm a^{2}}}={\frac {1}{2}}\ln \left|x^{2}\pm a^{2}\right|}
∫
x
d
x
(
x
2
±
a
2
)
2
=
−
1
2
(
x
2
±
a
2
)
{\displaystyle \int {\frac {x\;\mathrm {d} x}{(x^{2}\pm a^{2})^{2}}}=-{\frac {1}{2(x^{2}\pm a^{2})}}}
∫
x
d
x
(
x
2
±
a
2
)
3
=
−
1
4
(
x
2
±
a
2
)
2
{\displaystyle \int {\frac {x\;\mathrm {d} x}{(x^{2}\pm a^{2})^{3}}}=-{\frac {1}{4(x^{2}\pm a^{2})^{2}}}}
∫
x
d
x
(
x
2
±
a
2
)
n
=
−
1
2
(
n
−
1
)
(
x
2
±
a
2
)
n
−
1
(
n
≠
1
)
{\displaystyle \int {\frac {x\;\mathrm {d} x}{(x^{2}\pm a^{2})^{n}}}=-{\frac {1}{2(n-1)(x^{2}\pm a^{2})^{n-1}}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}
∫
x
2
d
x
x
2
+
a
2
=
x
−
a
a
r
c
t
a
n
x
a
{\displaystyle \int {\frac {x^{2}\mathrm {d} x}{x^{2}+a^{2}}}=x-a\,\mathrm {arctan} {\frac {x}{a}}}
∫
x
2
d
x
x
2
−
a
2
=
x
−
a
2
ln
|
x
+
a
x
−
a
|
=
x
−
{
a
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
a
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&=x-{\begin{cases}a\,\mathrm {artanh} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|<|a|\\a\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
x
2
d
x
(
x
2
+
a
2
)
2
=
−
x
2
(
x
2
+
a
2
)
+
1
2
a
a
r
c
t
a
n
x
a
{\displaystyle \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}}}
∫
x
2
d
x
(
x
2
−
a
2
)
2
=
−
x
2
(
x
2
−
a
2
)
−
1
4
a
ln
|
x
+
a
x
−
a
|
=
−
x
2
(
x
2
−
a
2
)
−
{
1
2
a
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
1
2
a
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&={\frac {-x}{2(x^{2}-a^{2})}}-{\begin{cases}{\frac {1}{2a}}\,\mathrm {artanh} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{2a}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
x
2
d
x
(
x
2
+
a
2
)
3
=
−
x
4
(
x
2
+
a
2
)
2
+
x
8
a
2
(
x
2
+
a
2
)
+
1
8
a
3
a
r
c
t
a
n
x
a
{\displaystyle \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}}}
∫
x
2
d
x
(
x
2
−
a
2
)
3
=
−
x
4
(
x
2
−
a
2
)
2
−
x
8
a
2
(
x
2
−
a
2
)
+
1
16
a
3
ln
|
x
+
a
x
−
a
|
=
−
x
4
(
x
2
−
a
2
)
2
−
x
8
a
2
(
x
2
−
a
2
)
+
{
1
8
a
3
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
1
8
a
3
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&={\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}}\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{8a^{3}}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
Rekursionsformel:
∫
x
2
d
x
(
x
2
±
a
2
)
n
=
−
x
2
(
n
−
1
)
(
x
2
±
a
2
)
n
−
1
+
1
2
(
n
−
1
)
∫
d
x
(
x
2
±
a
2
)
n
−
1
(
n
≠
1
)
{\displaystyle \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{)}}}
∫
x
3
d
x
x
2
±
a
2
=
x
2
2
∓
a
2
2
ln
|
x
2
±
a
2
|
{\displaystyle \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|}
∫
x
3
d
x
(
x
2
±
a
2
)
2
=
±
a
2
2
(
x
2
±
a
2
)
+
1
2
ln
|
x
2
±
a
2
|
{\displaystyle \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|}
∫
x
3
d
x
(
x
2
±
a
2
)
3
=
−
1
2
(
x
2
±
a
2
)
±
a
2
4
(
x
2
±
a
2
)
2
{\displaystyle \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}}}}
∫
x
3
d
x
(
x
2
±
a
2
)
n
=
−
1
2
(
n
−
2
)
(
x
2
±
a
2
)
n
−
2
±
a
2
2
(
n
−
1
)
(
x
2
±
a
2
)
n
−
1
(
n
>
2
)
{\displaystyle \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}}}\qquad {\mbox{(}}n>2{\mbox{)}}}
∫
d
x
x
(
x
2
±
a
2
)
=
±
1
2
a
2
ln
|
x
2
x
2
±
a
2
|
{\displaystyle \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|}
∫
d
x
x
(
x
2
±
a
2
)
2
=
±
1
2
a
2
(
x
2
±
a
2
)
+
1
2
a
4
ln
|
x
2
x
2
±
a
2
|
{\displaystyle \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|}
∫
d
x
x
(
x
2
±
a
2
)
3
=
±
1
4
a
2
(
x
2
±
a
2
)
2
+
1
2
a
4
(
x
2
±
a
2
)
±
1
2
a
6
ln
|
x
2
x
2
±
a
2
|
{\displaystyle \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|}
∫
d
x
x
2
(
x
2
+
a
2
)
=
−
1
a
2
x
−
1
a
3
a
r
c
t
a
n
x
a
{\displaystyle \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}}}
∫
d
x
x
2
(
x
2
−
a
2
)
=
1
a
2
x
−
1
2
a
3
ln
|
x
+
a
x
−
a
|
=
1
a
2
x
−
{
1
a
3
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
1
a
3
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&={\frac {1}{a^{2}x}}-{\begin{cases}{\frac {1}{a^{3}}}\,\mathrm {artanh} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|<|a|\\{\frac {1}{a^{3}}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
d
x
x
2
(
x
2
+
a
2
)
2
=
−
1
a
4
x
−
x
2
a
4
(
x
2
+
a
2
)
−
3
2
a
5
a
r
c
t
a
n
x
a
{\displaystyle \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}}}
∫
d
x
x
2
(
x
2
−
a
2
)
2
=
−
1
a
4
x
−
x
2
a
4
(
x
2
−
a
2
)
+
3
4
a
5
ln
|
x
+
a
x
−
a
|
=
−
1
a
4
x
−
x
2
a
4
(
x
2
−
a
2
)
+
{
3
2
a
5
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
3
2
a
5
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&={\frac {-1}{a^{4}x}}-{\frac {x}{2a^{4}(x^{2}-a^{2})}}+{\begin{cases}{\frac {3}{2a^{5}}}\,\mathrm {artanh} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|<|a|\\{\frac {3}{2a^{5}}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
d
x
x
2
(
x
2
+
a
2
)
3
=
−
1
a
6
x
−
x
4
a
4
(
x
2
+
a
2
)
2
−
7
x
8
a
6
(
x
2
+
a
2
)
−
15
8
a
7
a
r
c
t
a
n
x
a
{\displaystyle \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}}}
∫
d
x
x
2
(
x
2
−
a
2
)
3
=
1
a
6
x
−
x
4
a
4
(
x
2
−
a
2
)
2
+
7
x
8
a
6
(
x
2
−
a
2
)
−
15
16
a
7
ln
|
x
+
a
x
−
a
|
=
1
a
6
x
−
x
4
a
4
(
x
2
−
a
2
)
2
+
7
x
8
a
6
(
x
2
−
a
2
)
−
{
15
8
a
7
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
15
8
a
7
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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|\\&={\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}}\qquad {\text{wenn }}|x|<|a|\\{\frac {15}{8a^{7}}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
∫
d
x
x
3
(
x
2
±
a
2
)
=
∓
1
2
a
2
x
2
−
1
2
a
4
ln
|
x
2
x
2
±
a
2
|
{\displaystyle \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|}
∫
d
x
x
3
(
x
2
±
a
2
)
2
=
−
1
2
a
4
x
2
−
1
2
a
4
(
x
2
±
a
2
)
∓
1
a
6
ln
|
x
2
x
2
±
a
2
|
{\displaystyle \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|}
∫
d
x
x
3
(
x
2
±
a
2
)
3
=
∓
1
2
a
6
x
2
∓
1
a
6
(
x
2
±
a
2
)
−
1
4
a
4
(
x
2
±
a
2
)
2
−
3
2
a
8
ln
|
x
2
x
2
±
a
2
|
{\displaystyle \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|}
Integrale, die x2 ± a2 und einen Linearfaktor enthalten
Bearbeiten
∫
d
x
(
x
2
+
a
2
)
(
b
x
+
c
)
=
1
a
2
b
2
+
c
2
(
b
2
ln
(
b
x
+
c
)
2
x
2
+
a
2
+
c
a
arctan
x
a
)
{\displaystyle \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)}
∫
d
x
(
x
2
−
a
2
)
(
b
x
+
c
)
=
1
c
2
−
a
2
b
2
(
b
2
ln
|
(
b
x
+
c
)
2
x
2
−
a
2
|
−
c
2
a
ln
|
x
+
a
x
−
a
|
)
=
b
2
(
c
2
−
a
2
b
2
)
ln
|
(
b
x
+
c
)
2
x
2
−
a
2
|
−
{
c
a
(
c
2
−
a
2
b
2
)
a
r
t
a
n
h
x
a
wenn
|
x
|
<
|
a
|
c
a
(
c
2
−
a
2
b
2
)
a
r
c
o
t
h
x
a
wenn
|
x
|
>
|
a
|
{\displaystyle {\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)\\&={\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}}\qquad {\text{wenn }}|x|<|a|\\{\frac {c}{a(c^{2}-a^{2}b^{2})}}\,\mathrm {arcoth} \,{\frac {x}{a}}\qquad {\text{wenn }}|x|>|a|\end{cases}}\end{aligned}}}
Integrale, die ax2 + bx + c enthalten
Bearbeiten
Grundlage bildet das Integral
∫
d
x
a
x
2
+
b
x
+
c
=
{
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
+
C
(
4
a
c
−
b
2
>
0
)
1
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
+
C
(
4
a
c
−
b
2
<
0
)
{\displaystyle \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
4
a
c
−
b
2
<
0
{\displaystyle 4ac-b^{2}<0}
∫
d
x
a
x
2
+
b
x
+
c
=
−
2
b
2
−
4
a
c
⋅
{
a
r
t
a
n
h
2
a
x
+
b
b
2
−
4
a
c
+
C
wenn
|
2
a
x
+
b
b
2
−
4
a
c
|
<
1
a
r
c
o
t
h
2
a
x
+
b
b
2
−
4
a
c
+
C
wenn
|
2
a
x
+
b
b
2
−
4
a
c
|
>
1
{\displaystyle \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}}}
∫
d
x
(
a
x
2
+
b
x
+
c
)
2
=
2
a
x
+
b
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
+
2
a
4
a
c
−
b
2
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
∫
d
x
(
a
x
2
+
b
x
+
c
)
3
=
2
a
x
+
b
4
a
c
−
b
2
(
1
2
(
a
x
2
+
b
x
+
c
)
2
+
3
a
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
)
+
6
a
2
(
4
a
c
−
b
2
)
2
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
Rekursionsformel:
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
=
2
a
x
+
b
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
+
2
(
2
n
−
3
)
a
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
(
n
>
1
)
{\displaystyle \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}}}\qquad {\mbox{(}}n>1{\mbox{)}}}
∫
x
d
x
a
x
2
+
b
x
+
c
=
1
2
a
ln
|
a
x
2
+
b
x
+
c
|
−
b
2
a
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
In geschlossener Form ist
∫
e
x
+
f
a
x
2
+
b
x
+
c
d
x
=
e
2
a
ln
|
a
x
2
+
b
x
+
c
|
+
{
2
a
f
−
b
e
a
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
(
4
a
c
−
b
2
>
0
)
2
a
f
−
b
e
2
a
b
2
−
4
a
c
ln
|
2
a
x
+
b
−
b
2
−
4
a
c
2
a
x
+
b
+
b
2
−
4
a
c
|
+
C
(
4
a
c
−
b
2
<
0
)
{\displaystyle \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}}}}\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}}}
artanh und arcoth analog zu oben)
∫
x
d
x
(
a
x
2
+
b
x
+
c
)
2
=
−
b
x
+
2
c
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
−
b
4
a
c
−
b
2
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
Rekursionsformel:
∫
x
d
x
(
a
x
2
+
b
x
+
c
)
n
=
−
b
x
+
2
c
(
n
−
1
)
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
n
−
1
−
(
2
n
−
3
)
b
(
n
−
1
)
(
4
a
c
−
b
2
)
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
(
n
>
1
)
{\displaystyle \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}}}\qquad {\mbox{(}}n>1{\mbox{)}}}
∫
x
2
d
x
a
x
2
+
b
x
+
c
=
x
a
−
b
2
a
2
ln
|
a
x
2
+
b
x
+
c
|
+
b
2
−
2
a
c
2
a
2
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
∫
x
2
d
x
(
a
x
2
+
b
x
+
c
)
2
=
(
b
2
−
2
a
c
)
x
+
b
c
a
(
4
a
c
−
b
2
)
(
a
x
2
+
b
x
+
c
)
+
2
c
4
a
c
−
b
2
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
Die folgende Formel gilt für alle ganzen Zahlen n. Für n = 1 und n = 2 handelt es sich um die eben genannten Stammfunktionen:
∫
x
2
d
x
(
a
x
2
+
b
x
+
c
)
n
=
−
x
(
2
n
−
3
)
a
(
a
x
2
+
b
x
+
c
)
n
−
1
+
c
(
2
n
−
3
)
a
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
−
(
n
−
2
)
b
(
2
n
−
3
)
a
∫
x
d
x
(
a
x
2
+
b
x
+
c
)
n
{\displaystyle \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}}}}
Die folgende sehr allgemeine Formel gilt für alle ganzen Zahlen
m
{\displaystyle m}
n
≠
m
+
1
2
{\displaystyle n\neq {\frac {m+1}{2}}}
m
>
1
{\displaystyle m>1}
m
{\displaystyle m}
∫
x
m
d
x
(
a
x
2
+
b
x
+
c
)
n
=
−
x
m
−
1
(
2
n
−
m
−
1
)
a
(
a
x
2
+
b
x
+
c
)
n
−
1
+
(
m
−
1
)
c
(
2
n
−
m
−
1
)
a
∫
x
m
−
2
d
x
(
a
x
2
+
b
x
+
c
)
n
−
(
n
−
m
)
b
(
2
n
−
m
−
1
)
a
∫
x
m
−
1
d
x
(
a
x
2
+
b
x
+
c
)
n
(
m
≠
2
n
−
1
)
{\displaystyle {\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}}}\qquad {\mbox{(}}m\neq 2n-1{\mbox{)}}\end{aligned}}}
Für m = 2n–1 gilt statt dessen:
∫
x
2
n
−
1
d
x
(
a
x
2
+
b
x
+
c
)
n
=
1
a
∫
x
2
n
−
3
d
x
(
a
x
2
+
b
x
+
c
)
n
−
1
−
c
a
∫
x
2
n
−
3
d
x
(
a
x
2
+
b
x
+
c
)
n
−
b
a
∫
x
2
n
−
2
d
x
(
a
x
2
+
b
x
+
c
)
n
{\displaystyle \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}}}}
∫
d
x
x
(
a
x
2
+
b
x
+
c
)
=
1
2
c
ln
|
x
2
a
x
2
+
b
x
+
c
|
−
b
2
c
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
Rekursionsformel:
∫
d
x
x
(
a
x
2
+
b
x
+
c
)
n
=
1
2
c
(
n
−
1
)
(
a
x
2
+
b
x
+
c
)
n
−
1
−
b
2
c
∫
d
x
(
a
x
2
+
b
x
+
c
)
n
+
1
c
∫
d
x
x
(
a
x
2
+
b
x
+
c
)
n
−
1
+
C
(
n
>
1
)
{\displaystyle \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{)}}}
∫
d
x
x
2
(
a
x
2
+
b
x
+
c
)
=
b
2
c
2
ln
|
a
x
2
+
b
x
+
c
x
2
|
−
1
c
x
+
(
b
2
2
c
2
−
a
c
)
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
Und schließlich wieder als Rekursionsformel:
∫
d
x
x
m
(
a
x
2
+
b
x
+
c
)
n
=
−
1
(
m
−
1
)
c
x
m
−
1
(
a
x
2
+
b
x
+
c
)
n
−
1
−
(
2
n
+
m
−
3
)
a
(
m
−
1
)
c
∫
d
x
x
m
−
2
(
a
x
2
+
b
x
+
c
)
n
−
(
n
+
m
−
2
)
b
(
m
−
1
)
c
∫
d
x
x
m
−
1
(
a
x
2
+
b
x
+
c
)
n
(
m
>
1
)
{\displaystyle {\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}}}\qquad {\mbox{(}}m>1{\mbox{)}}\end{aligned}}}
Integrale, die ax2 + bx + c und einen Linearfaktor enthalten
Bearbeiten
∫
d
x
(
d
x
+
e
)
(
a
x
2
+
b
x
+
c
)
=
d
2
(
c
d
2
−
b
d
e
+
a
e
2
)
ln
|
(
d
x
+
e
)
2
a
x
2
+
b
x
+
c
|
+
2
a
e
−
b
d
2
(
c
d
2
−
b
d
e
+
a
e
2
)
∫
d
x
a
x
2
+
b
x
+
c
{\displaystyle \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}}}
Integrale, die drei oder mehr Linearfaktoren enthalten
Bearbeiten
Die Werte a, b, c,… müssen alle voneinander verschieden sein.
∫
d
x
(
x
+
a
)
(
x
+
b
)
(
x
+
c
)
=
ln
|
x
+
a
|
(
b
−
a
)
(
c
−
a
)
+
ln
|
x
+
b
|
(
a
−
b
)
(
c
−
b
)
+
ln
|
x
+
c
|
(
a
−
c
)
(
b
−
c
)
+
C
{\displaystyle \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}
∫
d
x
(
x
+
a
)
(
x
+
b
)
(
x
+
c
)
(
x
+
d
)
=
ln
|
x
+
a
|
(
b
−
a
)
(
c
−
a
)
(
d
−
a
)
+
ln
|
x
+
b
|
(
a
−
b
)
(
c
−
b
)
(
d
−
b
)
+
ln
|
x
+
c
|
(
a
−
c
)
(
b
−
c
)
(
d
−
c
)
+
ln
|
x
+
d
|
(
a
−
d
)
(
b
−
d
)
(
c
−
d
)
{\displaystyle {\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)}}\end{aligned}}}
…und so weiter. Allgemein geschrieben sieht das so aus:
∫
d
x
∏
i
=
1
n
(
x
+
a
i
)
=
∑
i
=
1
n
ln
|
x
+
a
i
|
∏
j
=
1
j
≠
i
n
(
a
j
−
a
i
)
(
∀
i
,
j
a
i
≠
a
j
)
{\displaystyle \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})}}\qquad {\mbox{(}}\forall _{i,j}\,a_{i}\neq a_{j}{\mbox{)}}}
Integrale, die x3 ± a3 enthalten
Bearbeiten
∫
d
x
x
3
±
a
3
=
1
6
a
2
ln
(
x
±
a
)
2
x
2
∓
a
x
+
a
2
±
1
a
2
3
arctan
2
x
∓
a
a
3
{\displaystyle \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}}}}}
∫
d
x
(
x
3
±
a
3
)
2
=
±
x
3
a
3
(
x
3
±
a
3
)
±
2
3
a
3
∫
d
x
x
3
±
a
3
{\displaystyle \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}}}}
∫
x
d
x
x
3
±
a
3
=
±
1
6
a
ln
x
2
∓
a
x
+
a
2
(
x
±
a
)
2
+
1
a
3
arctan
2
x
∓
a
a
3
{\displaystyle \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}}}}}
∫
x
d
x
(
x
3
±
a
3
)
2
=
±
x
2
3
a
3
(
x
3
±
a
3
)
±
1
3
a
3
∫
x
d
x
x
3
±
a
3
{\displaystyle \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}}}}
∫
x
2
d
x
x
3
±
a
3
=
±
1
3
ln
|
x
3
±
a
3
|
{\displaystyle \int {\frac {x^{2}\mathrm {d} x}{x^{3}\pm a^{3}}}=\pm {\frac {1}{3}}\ln |x^{3}\pm a^{3}|}
∫
x
2
d
x
(
x
3
±
a
3
)
2
=
−
1
3
(
x
3
±
a
3
)
{\displaystyle \int {\frac {x^{2}\mathrm {d} x}{(x^{3}\pm a^{3})^{2}}}={\frac {-1}{3(x^{3}\pm a^{3})}}}
∫
x
3
d
x
x
3
±
a
3
=
x
∓
a
3
∫
d
x
x
3
±
a
3
{\displaystyle \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}}}}
∫
x
3
d
x
(
x
3
±
a
3
)
2
=
−
x
3
(
x
3
±
a
3
)
+
1
3
∫
d
x
x
3
±
a
3
{\displaystyle \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}}}}
∫
d
x
x
(
x
3
±
a
3
)
=
±
1
3
a
3
ln
|
x
3
x
3
±
a
3
|
+
C
{\displaystyle \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}
∫
d
x
x
(
x
3
±
a
3
)
2
=
±
1
3
a
3
(
x
3
±
a
3
)
+
1
3
a
6
ln
|
x
3
x
3
±
a
3
|
{\displaystyle \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|}
∫
d
x
x
2
(
x
3
±
a
3
)
=
∓
1
a
3
x
∓
1
a
3
∫
x
d
x
x
3
±
a
3
+
C
{\displaystyle \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}
∫
d
x
x
2
(
x
3
±
a
3
)
2
=
−
1
a
6
x
−
x
2
3
a
6
(
x
3
±
a
3
)
−
4
3
a
6
∫
x
d
x
x
3
±
a
3
{\displaystyle \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}}}}
∫
d
x
x
3
(
x
3
±
a
3
)
=
∓
1
2
a
3
x
2
∓
1
a
3
∫
d
x
x
3
±
a
3
+
C
{\displaystyle \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}
∫
d
x
x
3
(
x
3
±
a
3
)
2
=
−
1
2
a
6
x
2
−
x
3
a
6
(
x
3
±
a
3
)
−
5
3
a
6
∫
d
x
x
3
±
a
3
{\displaystyle \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}}}}
Integrale, die x4 ± a4 enthalten
Bearbeiten
∫
d
x
x
4
+
a
4
=
1
4
2
a
3
(
ln
x
2
+
2
a
x
+
a
2
x
2
−
2
a
x
+
a
2
+
4
arctan
2
a
x
a
4
+
x
4
+
a
2
−
x
2
)
+
C
{\displaystyle \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}
∫
d
x
x
4
−
a
4
=
−
1
4
a
3
(
ln
|
x
+
a
x
−
a
|
+
2
arctan
x
a
)
+
C
{\displaystyle \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}
∫
x
d
x
x
4
+
a
4
=
1
2
a
2
arctan
x
2
a
2
+
C
{\displaystyle \int {\frac {x\,\mathrm {d} x}{x^{4}+a^{4}}}={\frac {1}{2a^{2}}}\arctan {\frac {x^{2}}{a^{2}}}+C}
∫
x
d
x
x
4
−
a
4
=
−
1
4
a
2
ln
|
x
2
+
a
2
x
2
−
a
2
|
+
C
{\displaystyle \int {\frac {x\,\mathrm {d} x}{x^{4}-a^{4}}}={\frac {-1}{4a^{2}}}\ln \left|{\frac {x^{2}+a^{2}}{x^{2}-a^{2}}}\right|+C}
∫
x
2
d
x
x
4
+
a
4
=
1
4
2
a
(
2
arctan
2
a
x
a
2
−
x
2
−
ln
x
2
+
2
a
x
+
a
2
x
2
+
2
a
x
+
a
2
)
+
C
{\displaystyle \int {\frac {x^{2}\mathrm {d} x}{x^{4}+a^{4}}}={\frac {1}{4{\sqrt {2}}\,a}}\left(2\arctan {\frac {{\sqrt {2}}\,ax}{a^{2}-x^{2}}}-\ln {\frac {x^{2}+{\sqrt {2}}\,ax+a^{2}}{x^{2}+{\sqrt {2}}\,ax+a^{2}}}\right)+C}
∫
x
2
d
x
x
4
−
a
4
=
1
4
a
(
2
arctan
x
a
−
ln
|
x
+
a
x
−
a
|
+
)
+
C
{\displaystyle \int {\frac {x^{2}\mathrm {d} x}{x^{4}-a^{4}}}={\frac {1}{4a}}\left(2\arctan {\frac {x}{a}}-\ln \left|{\frac {x+a}{x-a}}\right|+\right)+C}
∫
x
3
d
x
x
4
+
a
4
=
1
4
ln
(
x
4
+
a
4
)
+
C
{\displaystyle \int {\frac {x^{3}\mathrm {d} x}{x^{4}+a^{4}}}={\frac {1}{4}}\ln(x^{4}+a^{4})+C}
∫
x
3
d
x
x
4
−
a
4
=
1
4
ln
|
x
4
−
a
4
|
+
C
{\displaystyle \int {\frac {x^{3}\mathrm {d} x}{x^{4}-a^{4}}}={\frac {1}{4}}\ln \left|x^{4}-a^{4}\right|+C}
Integrale, die zwei quadratische Faktoren enthalten
Bearbeiten
∫
d
x
(
a
x
2
+
b
)
(
c
x
2
+
d
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{\displaystyle \int {\frac {\mathrm {d} x}{(ax^{2}+b)(cx^{2}+d)}}={\frac {1}{ad-bc}}\left(a\int {\frac {\mathrm {d} x}{ax^{2}+b}}-c\int {\frac {\mathrm {d} x}{cx^{2}+d}}\right)}
![{\displaystyle \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]\qquad {\mbox{(}}n\geq 1{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59fc6c782712d97c6ca8474ba10ab0992654aa74)
![{\displaystyle \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]\qquad {\mbox{(}}n\geq 1{\mbox{)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3565ae0799b25c54a6a50b9907a0c2a2152cb2d6)
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d17802daf33b5c8dffcc6260067b8b80586035da)
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f199c8a413580c8c53dd5537444d98e546aceea)
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd1ca3085b1f201acc52fcf4f95b46bd7d6f1d3e)
![{\displaystyle {\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]\qquad {\mbox{(}}n\geq 1{\mbox{)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1069ec5bf97f56e603b4493852f98fee324345a)
![{\displaystyle {\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]\\&{\mbox{(}}m\geq 1,\,n\geq 1{\mbox{)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f96ecc8322c0f3a3e1f568a8371c49bcdbd758f6)
