Calculus with the inverse hyperbolic functions

$\frac{d}{dx}\left(\operatorname{arsinh}x\right)=\frac{1}{\sqrt{x^2-1}}$

$\frac{d}{dx}\left(\operatorname{arsinh}x\right)=\frac{1}{\sqrt{1+x^2}}$

$\frac{d}{dx}\left(\operatorname{arcosh}x\right)=\frac{1}{\sqrt{x^2-1}}$

$\frac{d}{dx}\left(\operatorname{artanh}x\right)=\frac{1}{1-x^2}$

$\int_{ }^{ }\left(\frac{1}{\sqrt{x^2-a^2}}\right)dx=\operatorname{arcosh}\left(\frac{x}{a}\right)+C,\ x>a$

$\int_{ }^{ }\frac{1}{a^2-x^2}dx=\frac{1}{a}\operatorname{artanh}\left(\frac{x}{a}\right)+C,\ \left|x\right|<a$

$\int_{ }^{ }\left(\frac{1}{\sqrt{x^2+a^2}}\right)dx=\frac{1}{a}\operatorname{arsinh}\left(\frac{x}{a}\right)+C$

$\int_{ }^{ }\left(\frac{1}{\sqrt{x^2+a^2}}\right)dx=\operatorname{arsinh}\left(\frac{x}{a}\right)+C$

 $\frac{2}{\sqrt{4x^2-1}}$  

 $\frac{1}{\sqrt{4x^2-1}}$  

 $\frac{2}{\sqrt{x^2-1}}$  

 $\frac{1}{\sqrt{x^2-1}}$  

 $\frac{1}{1-x^2}$  

 $\frac{1}{x^2-1}$  

 $\frac{1}{x^2\left(1-x^2\right)}$  

 $\frac{x^2}{x^2-1}$  

 $\frac{1}{4}\operatorname{arsinh}\left(\frac{5x}{4}\right)+C$  

 $\frac{1}{5}\operatorname{arsinh}\left(\frac{5x}{4}\right)+C$  

 $\frac{1}{4}\operatorname{arcosh}\left(\frac{5x}{4}\right)+C$  

 $\frac{1}{5}\operatorname{arsinh}\left(\frac{5x}{4}\right)+C$  

 $\ln\left(\frac{4+\sqrt{15}}{2+\sqrt{3}}\right)$  

 $\frac{1}{4}\ln\left(\frac{4-\sqrt{15}}{2-\sqrt{3}}\right)$  

 $\frac{1}{4}\ln\left(\frac{4+\sqrt{15}}{2+\sqrt{3}}\right)$  

 $\ln\left(\frac{4-\sqrt{15}}{2-\sqrt{3}}\right)$  

 $\frac{1}{\sqrt{2}}\operatorname{arsinh}\sqrt{2}\left(x+2\right)+C$  

 $\frac{1}{\sqrt{2}}\operatorname{arcosh}\sqrt{2}\left(x+2\right)+C$  

 $\frac{1}{2}\operatorname{arsinh}2\left(x+2\right)+C$  

 $\frac{1}{2}\operatorname{arcosh}2\left(x+2\right)+C$  

 $\operatorname{arcosh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C$  

 $\frac{1}{\sqrt{3}}\operatorname{arcosh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C$  

 $\operatorname{arsinh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C$  

 $\frac{1}{\sqrt{3}}\operatorname{arsinh}\left(\frac{\sqrt{3}\left(x-2\right)}{4}\right)+C$  

 $\frac{1}{4}\operatorname{artanh}\left(\frac{2x+3}{2}\right)+C$  

 $\frac{1}{4}\operatorname{artanh}\left(\frac{x+3}{2}\right)+C$  

 $\frac{1}{8}\operatorname{artanh}\left(\frac{x+3}{4}\right)+C$  

 $\frac{1}{8}\operatorname{artanh}\left(\frac{2x+3}{4}\right)+C$