structural:expansion
Standard integrals

∫ sinh (a x) d x = a − 1 cosh (a x) + C ∫ cosh (a x) d x = a − 1 sinh (a x) + C ∫ tanh (a x) d x = a − 1 ln (cosh (a x)) + C ∫ coth (a x) d x = a − 1 ln | sinh (a x) | + C ∫ sech (a x) d x = a − 1 arctan (sinh (a x)) + C ∫ csch (a x) d x = a − 1 ln | tanh (a x 2) | + C = a − 1 ln | coth (a x) − csch (a x) | + C = − a − 1 arcoth (cosh (a x)) + C {\displaystyle {\begin{aligned}\int \sinh(ax)\,dx&=a^{-1}\cosh(ax)+C\\\int \cosh(ax)\,dx&=a^{-1}\sinh(ax)+C\\\int \tanh(ax)\,dx&=a^{-1}\ln(\cosh(ax))+C\\\int \coth(ax)\,dx&=a^{-1}\ln \left|\sinh(ax)\right|+C\\\int \operatorname {sech} (ax)\,dx&=a^{-1}\arctan(\sinh(ax))+C\\\int \operatorname {csch} (ax)\,dx&=a^{-1}\ln \left|\tanh \left({\frac {ax}{2}}\right)\right|+C=a^{-1}\ln \left|\coth \left(ax\right)-\operatorname {csch} \left(ax\right)\right|+C=-a^{-1}\operatorname {arcoth} \left(\cosh \left(ax\right)\right)+C\end{aligned}}}

The following integrals can be proved using hyperbolic substitution: ∫ 1 a 2 + u 2 d u = arsinh (u a) + C ∫ 1 u 2 − a 2 d u = sgn u arcosh | u a | + C ∫ 1 a 2 − u 2 d u = a − 1 artanh (u a) + C u 2 < a 2 ∫ 1 a 2 − u 2 d u = a − 1 arcoth (u a) + C u 2 > a 2 ∫ 1 u a 2 − u 2 d u = − a − 1 arsech | u a | + C ∫ 1 u a 2 + u 2 d u = − a − 1 arcsch | u a | + C {\displaystyle {\begin{aligned}\int {{\frac {1}{\sqrt {a^{2}+u^{2}}}}\,du}&=\operatorname {arsinh} \left({\frac {u}{a}}\right)+C\\\int {{\frac {1}{\sqrt {u^{2}-a^{2}}}}\,du}&=\operatorname {sgn} {u}\operatorname {arcosh} \left|{\frac {u}{a}}\right|+C\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {artanh} \left({\frac {u}{a}}\right)+C&&u^{2}<a^{2}\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {arcoth} \left({\frac {u}{a}}\right)+C&&u^{2}>a^{2}\\\int {{\frac {1}{u{\sqrt {a^{2}-u^{2}}}}}\,du}&=-a^{-1}\operatorname {arsech} \left|{\frac {u}{a}}\right|+C\\\int {{\frac {1}{u{\sqrt {a^{2}+u^{2}}}}}\,du}&=-a^{-1}\operatorname {arcsch} \left|{\frac {u}{a}}\right|+C\end{aligned}}}

where C is the constant of integration.