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Ein allgemeines Lösungsverfahren, wie es für rationale Funktionen vorliegt, gibt es für die Integrale algebraischer Funktionen nicht.

${\displaystyle \int x^{q}\;\mathrm {d} x={\frac {x^{q+1}}{q+1}}+C\qquad \left(q\neq -1\right)}$
ist Grundintegral auch für ${\displaystyle q\in \mathbb {Q} }$, ja sogar für ${\displaystyle q\in \mathbb {R} }$.

In verschiedenen Fällen hilfreich ist
${\displaystyle \int x^{n}(ax+b)^{q}\mathrm {d} x={\frac {1}{a^{n+1}}}\int {\left(X-b\right)^{n}X^{q}\mathrm {d} X}{\text{ mit }}X=ax+b\qquad {\mbox{(}}q\not \in \{-1,-2,\ldots ,-n\}{\mbox{)}},}$
denn für ${\displaystyle n\in \mathbb {N} }$ kann ${\displaystyle \left(X-b\right)^{n}}$ nach dem binomischen Lehrsatz entwickelt und dann das Grundintegral angewandt werden.

Die folgenden Tabellen enthalten ausschließlich Integrale von Funktionen, die Wurzelausdrücke enthalten. Mit Ausnahme des letzten Abschnitts kommen sogar nur Quadratwurzeln vor.

Integrale, die √x enthalten

Es wird ${\displaystyle a>0}$ vorausgesetzt, was für die hier behandelten Integrale keine Einschränkung bedeutet.

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

${\displaystyle \int {\frac {\mathrm {d} x}{{\sqrt {x}}\;(x+a^{2})}}={\frac {2}{a}}\arctan {\frac {\sqrt {x}}{a}}+C}$

${\displaystyle \int {\frac {x{\sqrt {x}}\;\mathrm {d} x}{x+a^{2}}}=2{\sqrt {x}}\left({\frac {x}{3}}-a^{2}\right)+2a^{3}\arctan {\frac {\sqrt {x}}{a}}+C}$

${\displaystyle \int {\frac {\mathrm {d} x}{x{\sqrt {x}}\;(x+a^{2})}}={\frac {-2}{a^{3}}}\left({\frac {a}{\sqrt {x}}}+\arctan {\frac {\sqrt {x}}{a}}\right)+C}$

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

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

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

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

• ${\displaystyle \int {\frac {{\sqrt {x}}\;\mathrm {d} x}{x^{2}+a^{2}}}={\frac {1}{2{\sqrt {2a}}}}\left(2\operatorname {arccot} {\frac {a-x}{\sqrt {2ax}}}+\ln {\frac {x-{\sqrt {2ax}}+a}{x+{\sqrt {2ax}}+a}}\right)+{\begin{cases}C&{\text{wenn }}x\leq a\\{\frac {\pi }{\sqrt {2a}}}+C&{\text{wenn }}x>a\end{cases}}}$

${\displaystyle \int {\frac {\mathrm {d} x}{{\sqrt {x}}(x^{2}+a^{2})}}={\frac {1}{2a{\sqrt {2a}}}}\left(2\operatorname {arccot} {\frac {a-x}{\sqrt {2ax}}}+\ln {\frac {x+{\sqrt {2ax}}+a}{x-{\sqrt {2ax}}+a}}\right)+{\begin{cases}C&{\text{wenn }}x\leq a\\{\frac {\pi }{a{\sqrt {2a}}}}+C&{\text{wenn }}x>a\end{cases}}}$

Die Änderung der Integrationskonstanten ist nötig, um eine Sprungstelle bei x = a zu vermeiden.

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

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

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

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

${\displaystyle \int {\frac {x{\sqrt {x}}\;\mathrm {d} x}{(x+a^{2})^{2}}}={\frac {{\sqrt {x}}\;(3a^{2}+2x)}{x+a^{2}}}-3a\arctan {\frac {\sqrt {x}}{a}}+C}$

${\displaystyle \int {\frac {\mathrm {d} x}{x{\sqrt {x}}\;(x+a^{2})^{2}}}={\frac {-1}{a^{2}(x+a^{2})}}\left({\frac {2}{\sqrt {x}}}+{\frac {3{\sqrt {x}}}{a^{2}}}\right)-{\frac {3}{a^{5}}}\arctan {\frac {\sqrt {x}}{a}}+C}$

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

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

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

${\displaystyle \int {\frac {\mathrm {d} x}{x{\sqrt {x}}\;(x-a^{2})^{2}}}={\frac {1}{a^{2}(x-a^{2})}}\left({\frac {2}{\sqrt {x}}}-{\frac {3{\sqrt {x}}}{a^{2}}}\right)+{\frac {3}{2a^{5}}}\ln \left|{\frac {{\sqrt {x}}+a}{{\sqrt {x}}-a}}\right|+C}$

Integrale, die √(ax + b) enthalten

• ${\displaystyle \int {\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{3a}}(ax+b){\sqrt {ax+b}}+C}$

${\displaystyle \int x{\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{15a^{2}}}(3ax-2b)(ax+b){\sqrt {ax+b}}+C}$

${\displaystyle \int x^{2}{\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{105a^{3}}}(15a^{2}x^{2}-12abx+8b^{2})(ax+b){\sqrt {ax+b}}+C}$

Rekursionsformel:
${\displaystyle \int x^{n}{\sqrt {ax+b}}\,\mathrm {d} x\;=\;{\frac {2}{a(2n+3)}}\left(x^{n}(ax+b){\sqrt {ax+b}}-nb\int x^{n-1}{\sqrt {ax+b}}\,\mathrm {d} x\right)\qquad {\mbox{(}}n\,\geq \,1{\mbox{)}}}$

• ${\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {ax+b}}}\,=\,{\frac {2}{a}}{\sqrt {ax+b}}+C}$

${\displaystyle \int {\frac {x\;\mathrm {d} x}{\sqrt {ax+b}}}\,=\,{\frac {2}{3a^{2}}}(ax-2b){\sqrt {ax+b}}+C}$

${\displaystyle \int {\frac {x^{2}\;\mathrm {d} x}{\sqrt {ax+b}}}\,=\,{\frac {2}{15a^{3}}}(3a^{2}x^{2}-4abx+8b^{2}){\sqrt {ax+b}}+C}$

Rekursionsformel:
${\displaystyle \int {\frac {x^{n}\,\mathrm {d} x}{\sqrt {ax+b}}}\;=\;{\frac {2}{a(2n+1)}}\left(x^{n}{\sqrt {ax+b}}-bn\int {\frac {x^{n-1}}{\sqrt {ax+b}}}\right)\qquad {\mbox{(}}n\,\geq \,1{\mbox{)}}}$

• ${\displaystyle \int {\frac {\mathrm {d} x}{x\;{\sqrt {ax+b}}}}\,=\,{\begin{cases}{\frac {1}{\sqrt {b}}}\ln \left|{\frac {{\sqrt {ax+b}}-{\sqrt {b}}}{{\sqrt {ax+b}}+{\sqrt {b}}}}\right|+C\qquad {\mbox{(}}b\,>\,0{\mbox{)}}\\{\frac {2}{\sqrt {-b}}}\arctan {\sqrt {\frac {ax+b}{-b}}}+C\qquad {\mbox{(}}b\,<\,0{\mbox{)}}\end{cases}}}$

Für ${\displaystyle b>0}$ kann man dieselbe Stammfunktion auch schreiben als
${\displaystyle \int {\frac {\mathrm {d} x}{x\;{\sqrt {ax+b}}}}\,=\,{\frac {-2}{\sqrt {b}}}\;\mathrm {artanh} {\sqrt {\frac {ax+b}{b}}}+C}$
Allerdings ist diese Funktion nur für ${\displaystyle \operatorname {sgn} x\neq \operatorname {sgn} a}$ definiert.

Viele der folgenden Integrale werden auf dieses Integral zurückgeführt. Die Fallunterscheidung ist dann wie oben vorzunehmen. Auch für die Schreibung mit artanh gilt das eben Gesagte.

${\displaystyle \int {\frac {\mathrm {d} x}{x^{2}{\sqrt {ax+b}}}}\,=\,{\frac {-{\sqrt {ax+b}}}{bx}}-{\frac {a}{2b}}\int {\frac {\mathrm {d} x}{x{\sqrt {ax+b}}}}}$

Rekursionsformel:
${\displaystyle \int {\frac {\mathrm {d} x}{x^{n}{\sqrt {ax+b}}}}\,=\,{\frac {-1}{b(n-1)}}\left({\frac {\sqrt {ax+b}}{x^{n-1}}}+{\frac {2n-3}{2}}\;\cdot a\int {\frac {\mathrm {d} x}{x^{n-1}{\sqrt {ax+b}}}}\right)\qquad {\mbox{(}}n\,\geq \,2{\mbox{)}}}$

• ${\displaystyle \int {\frac {\sqrt {ax+b}}{x}}\;\mathrm {d} x\,=\,2{\sqrt {ax+b}}+b\int {\frac {\mathrm {d} x}{x{\sqrt {ax+b}}}}}$

${\displaystyle \int {\frac {\sqrt {ax+b}}{x^{2}}}\;\mathrm {d} x\,=\,{\frac {-{\sqrt {ax+b}}}{x}}+{\frac {a}{2}}\int {\frac {\mathrm {d} x}{x{\sqrt {ax+b}}}}}$

• ${\displaystyle \int (ax+b){\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{5a}}(ax+b)^{2}{\sqrt {ax+b}}+C}$

${\displaystyle \int x(ax+b){\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{35a^{2}}}(5ax-2b)(ax+b)^{2}{\sqrt {ax+b}}+C}$

${\displaystyle \int x^{2}(ax+b){\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{315a^{3}}}(35a^{2}x^{2}-20abx+8b^{2})(ax+b)^{2}{\sqrt {ax+b}}+C}$

${\displaystyle \int {\frac {1}{x}}(ax+b){\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{3}}(ax+4b){\sqrt {ax+b}}+b^{2}\int {\frac {\mathrm {d} x}{x{\sqrt {ax+b}}}}}$

• ${\displaystyle \int {\frac {\mathrm {d} x}{(ax+b){\sqrt {ax+b}}}}\,=\,{\frac {-2}{a{\sqrt {ax+b}}}}+C}$

${\displaystyle \int {\frac {x\;\mathrm {d} x}{(ax+b){\sqrt {ax+b}}}}\,=\,{\frac {2(ax+2b)}{a^{2}{\sqrt {ax+b}}}}+C}$

${\displaystyle \int {\frac {x^{2}\mathrm {d} x}{(ax+b){\sqrt {ax+b}}}}\,=\,{\frac {2(a^{2}x^{2}-4abx-8b^{2})}{3a^{3}{\sqrt {ax+b}}}}+C}$

${\displaystyle \int {\frac {\mathrm {d} x}{x(ax+b){\sqrt {ax+b}}}}\,=\,{\frac {1}{b}}\left({\frac {2}{a{\sqrt {ax+b}}}}+\int {\frac {\mathrm {d} x}{x{\sqrt {ax+b}}}}\right)}$

${\displaystyle \int {\frac {\mathrm {d} x}{x^{2}(ax+b){\sqrt {ax+b}}}}\,=\,{\frac {-1}{b^{2}}}\left({\frac {3ax+b}{x{\sqrt {ax+b}}}}+{\frac {3}{2a}}\int {\frac {\mathrm {d} x}{x{\sqrt {ax+b}}}}\right)}$

• Die jetzt noch folgenden Integrale sind für n = 0, n = ±1 und zum Teil für n = –2 oben bereits behandelt worden. Diese Formeln lassen sich aber verallgemeinern:
  ${\displaystyle \int (ax+b)^{n}{\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{a(2n+3)}}(ax+b)^{n+1}{\sqrt {ax+b}}+C\qquad {\mbox{(}}n\,\in \,\mathbb {Z} {\mbox{)}}}$

${\displaystyle \int x(ax+b)^{n}{\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{a^{2}}}\cdot {\frac {(2n+3)ax-2b}{4n^{2}+16n+15}}(ax+b)^{n+1}{\sqrt {ax+b}}+C\qquad {\mbox{(}}n\,\in \,\mathbb {Z} {\mbox{)}}}$

${\displaystyle \int x^{2}(ax+b)^{n}{\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{a^{3}}}\cdot {\frac {(2n+3)(2n+5)a^{2}x^{2}-4(2n+3)abx+8b^{2}}{(2n+3)(2n+5)(2n+7)}}(ax+b)^{n+1}{\sqrt {ax+b}}+C\qquad {\mbox{(}}n\,\in \,\mathbb {Z} {\mbox{)}}}$

Auch die nächsten Formeln sind für n ∈ Z gültig. Als Rekursionsformeln taugen sie natürlich nur für n ≥ 1:
${\displaystyle \int {\frac {1}{x}}(ax+b)^{n}{\sqrt {ax+b}}\,\mathrm {d} x\,=\,{\frac {2}{2n+1}}(ax+b)^{n}{\sqrt {ax+b}}\;+\;b\int (ax+b)^{n-1}{\sqrt {ax+b}}\,\mathrm {d} x\qquad {\mbox{(}}n\,\geq \,1{\mbox{)}}}$

${\displaystyle \int {\frac {{\sqrt {ax+b}}\,\mathrm {d} x}{x(ax+b)^{n}}}\,=\,{\frac {1}{b}}\left({\frac {2}{2n-3}}{\frac {\sqrt {ax+b}}{(ax+b)^{n-1}}}\;+\;\int {\frac {\sqrt {ax+b}}{(ax+b)^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(}}n\,\geq \,1{\mbox{)}}}$

${\displaystyle \int {\frac {{\sqrt {ax+b}}\,\mathrm {d} x}{x^{2}(ax+b)^{n}}}\,=\,{\frac {-1}{b}}\left({\frac {\sqrt {ax+b}}{x(ax+b)^{n-1}}}\;+\;(2n-1)\;{\frac {a}{2}}\int {\frac {\sqrt {ax+b}}{(ax+b)^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(}}n\,\geq \,1{\mbox{)}}}$

Integrale die r = √(x² + a²) beinhalten

${\displaystyle \int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left(x+r\right)\right)}$

${\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {1}{8}}3a^{2}xr+{\frac {3}{8}}a^{4}\ln \left(x+r\right)}$

${\displaystyle \int r^{5}\;dx={\frac {1}{6}}xr^{5}+{\frac {5}{24}}a^{2}xr^{3}+{\frac {5}{16}}a^{4}xr+{\frac {5}{16}}a^{6}\ln \left(x+r\right)}$

${\displaystyle \int xr\;dx={\frac {r^{3}}{3}}}$

${\displaystyle \int xr^{3}\;dx={\frac {r^{5}}{5}}}$

${\displaystyle \int xr^{2n+1}\;dx={\frac {r^{2n+3}}{2n+3}}}$

${\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left(x+r\right)}$

${\displaystyle \int x^{2}r^{3}\;dx={\frac {xr^{5}}{6}}-{\frac {a^{2}xr^{3}}{24}}-{\frac {a^{4}xr}{16}}-{\frac {a^{6}}{16}}\ln \left(x+r\right)}$

${\displaystyle \int x^{3}r\;dx={\frac {r^{5}}{5}}-{\frac {a^{2}r^{3}}{3}}}$

${\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{7}}{7}}-{\frac {a^{2}r^{5}}{5}}}$

${\displaystyle \int x^{3}r^{2n+1}\;dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}}$

${\displaystyle \int x^{4}r\;dx={\frac {x^{3}r^{3}}{6}}-{\frac {a^{2}xr^{3}}{8}}+{\frac {a^{4}xr}{16}}+{\frac {a^{6}}{16}}\ln \left(x+r\right)}$

${\displaystyle \int x^{4}r^{3}\;dx={\frac {x^{3}r^{5}}{8}}-{\frac {a^{2}xr^{5}}{16}}+{\frac {a^{4}xr^{3}}{64}}+{\frac {3a^{6}xr}{128}}+{\frac {3a^{8}}{128}}\ln \left(x+r\right)}$

${\displaystyle \int x^{5}r\;dx={\frac {r^{7}}{7}}-{\frac {2a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}}$

${\displaystyle \int x^{5}r^{3}\;dx={\frac {r^{9}}{9}}-{\frac {2a^{2}r^{7}}{7}}+{\frac {a^{4}r^{5}}{5}}}$

${\displaystyle \int x^{5}r^{2n+1}\;dx={\frac {r^{2n+7}}{2n+7}}-{\frac {2a^{2}r^{2n+5}}{2n+5}}+{\frac {a^{4}r^{2n+3}}{2n+3}}}$

${\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\sinh ^{-1}{\frac {a}{x}}}$

${\displaystyle \int {\frac {r^{3}\;dx}{x}}={\frac {r^{3}}{3}}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|}$

${\displaystyle \int {\frac {r^{5}\;dx}{x}}={\frac {r^{5}}{5}}+{\frac {a^{2}r^{3}}{3}}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}}\right|}$

${\displaystyle \int {\frac {r^{7}\;dx}{x}}={\frac {r^{7}}{7}}+{\frac {a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}+a^{6}r-a^{7}\ln \left|{\frac {a+r}{x}}\right|}$

${\displaystyle \int {\frac {dx}{r}}=\sinh ^{-1}{\frac {x}{a}}=\ln \left|x+r\right|}$

${\displaystyle \int {\frac {dx}{r^{3}}}={\frac {x}{a^{2}r}}}$

${\displaystyle \int {\frac {x\,dx}{r}}=r}$

${\displaystyle \int {\frac {x\,dx}{r^{3}}}=-{\frac {1}{r}}}$

${\displaystyle \int {\frac {x^{2}\;dx}{r}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\,\sinh ^{-1}{\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left|x+r\right|}$

${\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\sinh ^{-1}{\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|}$

Integrale die s = √(x² - a²) beinhalten

Annahme ${\displaystyle (x^{2}>a^{2})}$, für ${\displaystyle (x^{2}<a^{2})}$, siehe nächster Abschnitt:

${\displaystyle \int s\;dx={\frac {1}{2}}\left(xs-a^{2}\ln \left|{\frac {x+s}{a}}\right|\right)+C={\frac {1}{2}}\left(xs+a^{2}\ln \left|{\frac {x-s}{a}}\right|\right)+C={\frac {1}{2}}\left(xs-\operatorname {sgn} (x)\,a^{2}\operatorname {arcosh} \left|{\frac {x}{a}}\right|\right)+C}$

${\displaystyle \int xs\;dx={\frac {1}{3}}s^{3}}$

${\displaystyle \int {\frac {s\;dx}{x}}=s-|a|\arccos \left|{\frac {a}{x}}\right|}$

${\displaystyle \int {\frac {dx}{s}}=\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\ln \left|{\frac {x+s}{a}}\right|}$

Hierbei ist ${\displaystyle \ln \left|{\frac {x+s}{a}}\right|=\operatorname {sgn} (x)\,\operatorname {arcosh} \left|{\frac {x}{a}}\right|={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)}$, wobei der positive Wert des ${\displaystyle \operatorname {arcosh} \left|{\frac {x}{a}}\right|}$ genommen werden muss.

${\displaystyle \int {\frac {x\;dx}{s}}=s}$

${\displaystyle \int {\frac {x\;dx}{s^{3}}}=-{\frac {1}{s}}}$

${\displaystyle \int {\frac {x\;dx}{s^{5}}}=-{\frac {1}{3s^{3}}}}$

${\displaystyle \int {\frac {x\;dx}{s^{7}}}=-{\frac {1}{5s^{5}}}}$

${\displaystyle \int {\frac {x\;dx}{s^{2n+1}}}=-{\frac {1}{(2n-1)s^{2n-1}}}}$

${\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=-{\frac {1}{2n-1}}{\frac {x^{2m-1}}{s^{2n-1}}}+{\frac {2m-1}{2n-1}}\int {\frac {x^{2m-2}\;dx}{s^{2n-1}}}}$

${\displaystyle \int {\frac {x^{2}\;dx}{s}}={\frac {xs}{2}}+{\frac {a^{2}}{2}}\ln \left|{\frac {x+s}{a}}\right|}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{3}}}=-{\frac {x}{s}}+\ln \left|{\frac {x+s}{a}}\right|}$

${\displaystyle \int {\frac {x^{4}\;dx}{s}}={\frac {x^{3}s}{4}}+{\frac {3}{8}}a^{2}xs+{\frac {3}{8}}a^{4}\ln \left|{\frac {x+s}{a}}\right|}$

${\displaystyle \int {\frac {x^{4}\;dx}{s^{3}}}={\frac {xs}{2}}-{\frac {a^{2}x}{s}}+{\frac {3}{2}}a^{2}\ln \left|{\frac {x+s}{a}}\right|}$

${\displaystyle \int {\frac {x^{4}\;dx}{s^{5}}}=-{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}+\ln \left|{\frac {x+s}{a}}\right|}$

${\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=(-1)^{n-m}{\frac {1}{a^{2(n-m)}}}\sum _{i=0}^{n-m-1}{\frac {1}{2(m+i)+1}}{n-m-1 \choose i}{\frac {x^{2(m+i)+1}}{s^{2(m+i)+1}}}\qquad {\mbox{(}}n>m\geq 0{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{s^{3}}}=-{\frac {1}{a^{2}}}{\frac {x}{s}}}$

${\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}$

${\displaystyle \int {\frac {dx}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\frac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}$

${\displaystyle \int {\frac {dx}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\frac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{7}}}={\frac {1}{a^{4}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}$

${\displaystyle \int {\frac {x^{2}\;dx}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}$

Integrale die t = √(a² - x²) beinhalten

${\displaystyle \int t\;dx={\frac {1}{2}}\left(xt+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

${\displaystyle \int t\;dx={\frac {1}{2}}\left(xt-\operatorname {sgn} x\,\cosh ^{-1}\left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(}}|x|\geq |a|{\mbox{)}}}$

${\displaystyle \int xt\;dx=-{\frac {1}{3}}t^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

${\displaystyle \int x^{2}t\;dx={\frac {1}{8}}\left(2x^{3}t-xa^{2}t+a^{4}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

${\displaystyle \int {\frac {t\;dx}{x}}=t-a\ln \left|{\frac {a+t}{x}}\right|\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{t}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

${\displaystyle \int {\frac {x^{2}\;dx}{t}}=-{\frac {x}{2}}t+{\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

Integrale die √R = √(ax² + bx+ c) beinhalten

${\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {aR}}+2ax+b\right|\qquad {\mbox{(}}a>0{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\,\sinh ^{-1}{\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(}}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(}}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{\sqrt {ax^{2}+bx+c}}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(}}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{3}}}}={\frac {4ax+2b}{(4ac-b^{2}){\sqrt {R}}}}}$

${\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{5}}}}={\frac {4ax+2b}{3(4ac-b^{2}){\sqrt {R}}}}\left({\frac {1}{R}}+{\frac {8a}{4ac-b^{2}}}\right)}$

${\displaystyle \int {\frac {dx}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}={\frac {4ax+2b}{(2n-1)(4ac-b^{2})R^{(2n-1)/2}}}+{\frac {8a(n-1)}{(2n-1)(4ac-b^{2})}}\int {\frac {dx}{R^{(2n-1)/2}}}}$

${\displaystyle \int {\frac {x\;dx}{\sqrt {ax^{2}+bx+c}}}={\frac {\sqrt {R}}{a}}-{\frac {b}{2a}}\int {\frac {dx}{\sqrt {R}}}}$

${\displaystyle \int {\frac {x\;dx}{\sqrt {(ax^{2}+bx+c)^{3}}}}=-{\frac {2bx+4c}{(4ac-b^{2}){\sqrt {R}}}}}$

${\displaystyle \int {\frac {x\;dx}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}=-{\frac {1}{(2n-1)aR^{(2n-1)/2}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{(2n+1)/2}}}}$

${\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {cR}}+bx+2c}{x}}\right)}$

${\displaystyle \int {\frac {dx}{x{\sqrt {ax^{2}+bx+c}}}}=-{\frac {1}{\sqrt {c}}}\sinh ^{-1}\left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}$