Areaalfunctie

${\displaystyle \operatorname {arsinh} x=\ln \left(x+{\sqrt {x^{2}+1}}\right),\quad x\in \mathbb {R} }$

${\displaystyle \operatorname {arcosh} x=\ln \left(x+{\sqrt {x^{2}-1}}\right),\quad x\in \mathbb {R} ,\ x\geq 1}$

${\displaystyle \operatorname {artanh} x={\tfrac {1}{2}}\ln {\frac {1+x}{1-x}},\quad x\in \mathbb {R} ,\ |x|<1}$

${\displaystyle \operatorname {arcoth} x={\tfrac {1}{2}}\ln {\frac {x+1}{x-1}},\quad x\in \mathbb {R} ,\ |x|>1}$

${\displaystyle \operatorname {arsech} x=\ln \left({\frac {1}{x}}+{\sqrt {{\frac {1}{x^{2}}}-1}}\right),\quad x\in \mathbb {R} ,\ 0<x\leq 1}$

${\displaystyle \operatorname {arcsch} x=\ln \left({\frac {1}{x}}+{\sqrt {{\frac {1}{x^{2}}}+1}}\right),\quad x\in \mathbb {R} ,\ x\neq 0}$

${\displaystyle {\begin{aligned}2\cdot \operatorname {arcosh} x&=\operatorname {arcosh} (2x^{2}-1)&\quad {\hbox{ for }}x\geq 1\\4\cdot \operatorname {arcosh} x&=\operatorname {arcosh} (8x^{4}-8x^{2}+1)&\quad {\hbox{ for }}x\geq 1\\2\cdot \operatorname {arsinh} x&=\operatorname {arcosh} (2x^{2}+1)&\quad {\hbox{ for }}x\geq 0\\4\cdot \operatorname {arsinh} x&=\operatorname {arcosh} (8x^{4}+8x^{2}+1)&\quad {\hbox{ for }}x\geq 0\end{aligned}}}$

${\displaystyle \operatorname {arcosh} \left({\frac {x^{2}+1}{2x}}\right)=\operatorname {arsinh} \left({\frac {x^{2}-1}{2x}}\right)=\operatorname {artanh} \left({\frac {x^{2}-1}{x^{2}+1}}\right)=\ln(x)}$

${\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}\,\operatorname {arsinh} (x)\,={\frac {1}{\sqrt {x^{2}+1}}}}$

${\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}\,\operatorname {arcosh} (x)\,={\frac {1}{\sqrt {x^{2}-1}}}}$

${\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}\,\operatorname {artanh} (x)\,={\frac {1}{1-x^{2}}}}$

${\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}\,\operatorname {arcoth} (x)\,={\frac {-1}{-1+x^{2}}}}$

${\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}\,\operatorname {arsech} (x)\,=-{\frac {1}{x{\sqrt {1-x^{2}}}}}}$

${\displaystyle {\frac {\rm {d}}{{\rm {d}}x}}\,\operatorname {arcsch} (x)\,=-{\frac {1}{\left|x\right|{\sqrt {1+x^{2}}}}}}$

De integralen van de areaalfuncties kunnen alle berekend worden met partiële integratie.

${\displaystyle \int \operatorname {arsinh} (x){\rm {d}}x=x\ \operatorname {arsinh} (x)-{\sqrt {x^{2}+1}}+C}$

${\displaystyle \int \operatorname {arcosh} (x){\rm {d}}x=x\ \operatorname {arcosh} (x)-{\sqrt {x^{2}-1}}+C}$

${\displaystyle \int \operatorname {artanh} (x){\rm {d}}x=x\ \operatorname {artanh} (x)+{\tfrac {1}{2}}\ln(1-x^{2})+C}$

${\displaystyle \int \operatorname {arcoth} (x){\rm {d}}x=x\ \operatorname {arcoth} (x)+{\tfrac {1}{2}}\ln(x^{2}-1)+C}$

${\displaystyle \int \operatorname {arsech} (x){\rm {d}}x=x\ \operatorname {arsech} (x)-\operatorname {arctan} \left({\sqrt {{\frac {1}{x^{2}}}-1}}\right)+C}$

${\displaystyle \int \operatorname {arcsch} (x){\rm {d}}x=x\ \operatorname {arcsch} (x)+\ln \left(x+{\sqrt {x^{2}+1}}\right)+C}$

${\displaystyle \operatorname {arsinh} (x)+\operatorname {arsinh} (y)=\operatorname {arsinh} \left(x{\sqrt {1+y^{2}}}+y{\sqrt {1+x^{2}}}\right)}$

${\displaystyle \operatorname {arsinh} (x)-\operatorname {arsinh} (y)=\operatorname {arsinh} \left(x{\sqrt {1+y^{2}}}-y{\sqrt {1+x^{2}}}\right)}$

${\displaystyle \operatorname {arcosh} (x)+\operatorname {arcosh} (y)=\operatorname {arcosh} \left(xy+{\sqrt {x^{2}-1}}{\sqrt {y^{2}-1}}\right)}$

${\displaystyle \operatorname {arcosh} (x)-\operatorname {arcosh} (y)=\operatorname {arcosh} \left(xy-{\sqrt {x^{2}-1}}{\sqrt {y^{2}-1}}\right)}$

${\displaystyle \operatorname {artanh} (x)+\operatorname {artanh} (y)=\operatorname {artanh} \left({\frac {x+y}{1+xy}}\right)}$

${\displaystyle \operatorname {artanh} (x)-\operatorname {artanh} (y)=\operatorname {artanh} \left({\frac {x-y}{1-xy}}\right)}$

Bij het gebruik van deze uitdrukkingen moet rekening gehouden worden met de eventuele beperkingen op het domein van deze functies.

${\displaystyle \operatorname {arsinh} (x)=\pm \operatorname {arcosh} \left({\sqrt {x^{2}+1}}\right)=\operatorname {artanh} \left({\frac {x}{\sqrt {x^{2}+1}}}\right)}$

${\displaystyle \operatorname {arcosh} (x)=\operatorname {arsinh} \left({\sqrt {x^{2}-1}}\right)=\operatorname {artanh} \left({\frac {\sqrt {x^{2}-1}}{x}}\right)}$

${\displaystyle \operatorname {artanh} (x)=\operatorname {arsinh} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)=\pm \operatorname {arcosh} \left({\frac {1}{\sqrt {1-x^{2}}}}\right)=\operatorname {arcoth} \left({\frac {1}{x}}\right)}$

${\displaystyle \operatorname {arcoth} (x)=\operatorname {arsinh} \left({\frac {1}{\sqrt {x^{2}-1}}}\right)=\pm \operatorname {arcosh} \left({\frac {x}{\sqrt {x^{2}-1}}}\right)=\operatorname {artanh} \left({\frac {1}{x}}\right)}$

${\displaystyle \operatorname {arcsch} (x)=\operatorname {arsinh} \left({\frac {1}{x}}\right)}$

${\displaystyle \operatorname {arsech} (x)=\operatorname {arcosh} \left({\frac {1}{x}}\right)}$

Enkel de areaalsinus hyperbolicus en de areaaltangens hyperbolicus kunnen probleemloos rond x= 0 worden ontwikkeld. De reeksen zijn dan:

${\displaystyle \operatorname {arsinh} (x)=\sum _{n=0}^{\infty }\,(-1)^{n}\,{\frac {(2n)!}{2^{2n}(n!)^{2}}}\,{\frac {x^{2n+1}}{2n+1}},\quad |x|<1}$

${\displaystyle \operatorname {artanh} (x)=\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{2n+1}},\quad |x|<1}$

Een andere reeksontwikkeling is:

${\displaystyle \operatorname {arcosh} (x)=\ln(2x)-\sum _{n=1}^{\infty }{\frac {(2n)!}{2^{2n}(n!)^{2}}}{\frac {x^{-2n}}{2n}},\quad x>1}$

Samenstellingen van hyperbolische en areaalfuncties leveren irrationale functievoorschriften:

${\displaystyle \sinh(\operatorname {arcosh} (x))={\sqrt {x^{2}-1}}\quad$ :\quad x\geq 1}

${\displaystyle \sinh(\operatorname {artanh} (x))={\frac {x}{\sqrt {1-x^{2}}}}\quad$ :\quad -1<x<1}

${\displaystyle \cosh(\operatorname {arsinh} (x))={\sqrt {1+x^{2}}}}$

${\displaystyle \cosh(\operatorname {artanh} (x))={\frac {1}{\sqrt {1-x^{2}}}}\quad$ :\quad -1<x<1}

${\displaystyle \tanh(\operatorname {arsinh} (x))={\frac {x}{\sqrt {1+x^{2}}}}}$

${\displaystyle \tanh(\operatorname {arcosh} (x))={\frac {\sqrt {x^{2}-1}}{x}}\quad$ :\quad x\geq 1}

Twee standaardlimieten die areaalfuncties bevatten zijn:

${\displaystyle \lim _{x\to 0}\,{\frac {\operatorname {arsinh} (x)}{x}}=1}$

${\displaystyle \lim _{x\to 0}\,{\frac {\operatorname {artanh} (x)}{x}}=1}$

Beide volgen uit de soortgelijke limiet voor de overeenstemmende hyperbolische functie, door middel van een eenvoudige substitutie.