inverse hyperbolic functions

Differentiation

$$
y=\sinh^{-1}x\quad\Longrightarrow\quad x=\sinh y\\
\frac{{\rm d}x}{{\rm d}y}=\cosh y=\sqrt{(\sinh y)^2+1}=\sqrt{x^2+1}\\[6pt]
y=\cosh^{-1}x\quad\Longrightarrow\quad x=\cosh y\\
\frac{{\rm d}x}{{\rm d}y}=\sinh y=\sqrt{(\cosh y)^2-1}=\sqrt{x^2-1}\\[6pt]
$$

Integration

$$
\begin{aligned}
\int\sinh^{-1}x{\rm\ d}x&=x\sinh^{-1}x-\int x{\rm\ d}\sinh^{-1}x\\
&=x\sinh^{-1}x-\int\frac{x}{\sqrt{1+x^2}}{\rm\ d}x\\
&=x\sinh^{-1}x-\int\frac{1}{2}(1+x^2)^{-\frac{1}{2}}{\rm\ d}(1+x^2)\\
&=x\sinh^{-1}x-\sqrt{1+x^2}+c\\[6pt]
\int\cosh^{-1}x{\rm\ d}x&=x\cosh^{-1}x-\int x{\rm\ d}\cosh^{-1}x\\
&=x\cosh^{-1}x-\int\frac{x}{\sqrt{x^2-1}}{\rm\ d}x\\
&=x\cosh^{-1}x-\int\frac{1}{2}(x^2-1)^{-\frac{1}{2}}{\rm\ d}(x^2-1)\\
&=x\cosh^{-1}x-\sqrt{x^2-1}+c\\[6pt]
\int\tanh^{-1}x{\rm\ d}x&=x\tanh^{-1}x-\int x{\rm\ d}\tanh^{-1}x\\
&=x\tanh^{-1}x-\int\frac{x}{1-x^2}{\rm\ d}x\\
&=x\tanh^{-1}x+\frac{1}{2}\int\frac{1}{x^2-1}{\rm\ d}(x^2-1)\\
&=x\tanh^{-1}x+\frac{1}{2}\ln|x^2-1|+c\\[6pt]
\end{aligned}
$$