$$\begin{eqnarray*}\sinh^{-1}x&=&\log (x+\sqrt{x^{2}+1})\\\cosh^{-1}x&=&\log (x+\sqrt{x^{2}-1})\\\tanh^{-1}x&=&\frac{1}{2}\log\frac{1+x}{1-x}\end{eqnarray*} $$
Proof
Prove that $$\sinh^{-1}x=\log (x+\sqrt{x^{2}+1})$$
Let \(y=\sinh^{-1} x\). Then by the definition of inverse functions,
$$x=\sinh y=\frac{e^{y}-e^{-y}}{2}$$
so multiplying by \(e^{y}\), we have
$$e^{2y}-2xe^{y}-1=0$$
Considering it as a quadratic equation in \(e^{y}\), we see that
$$(e^{y})^{2}-2x(e^{y})-1=0$$
Solving by the quadratic formula, we get
$$e^{y}=\frac{2x\pm \sqrt{4x^{2}+4}}{2}=x\pm\sqrt{x^{2}+1}$$
Reminding that \(e^{y}>0\), but \(x-\sqrt{x^{2}+1}<0\) since \(x\leq \sqrt{x^{2}}<\sqrt{x^{2}+1} \).
Thus, we have
$$e^{y}=x+\sqrt{x^{2}+1}$$
Therefore, we get
$$y=\ln (e^{y})=\ln(x+\sqrt{x^{2}+1})$$
Similarly, we get \(\cosh^{-1}x=\log (x+\sqrt{x^{2}-1})\).
Prove that
$$\tanh^{-1}x=\frac{1}{2}\log\frac{1+x}{1-x}$$
Let \(y=\tanh^{-1} x\). Then by the definition of invers functions,
$$x=\tanh y=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}} \left(\times \frac{e^{y}}{e^{y}}\right)=\frac{e^{2y}-1}{e^{2y}+1}$$
so multiplying by \((e^{2y}+1)\), we have
$$x(e^{2y}+1)=e^{2y}-1$$
Considering it as a equation in \(e^{2y}\), we see that
$$(1-x)e^{2y}=1+x$$
Solving the equation about \(e^{2y}\), we get
$$e^{2y}=\frac{1+x}{1-x}$$
Therefore, we get
$$y=\frac{1}{2}\ln(e^{2y})=\frac{1}{2}\ln\frac{1+x}{1-x}$$