Answer
$$\tanh ^{-1} t=\frac{1}{2} \ln \left(\frac{1+t}{1-t}\right) \text { for }|t|<1$$
Work Step by Step
Let
$$ x=\tanh ^{1} t$$
Then
\begin{aligned}
t&=\tanh x \\
&=\frac{e^{x}-e^{-1}}{e^{x}+e^{-1}}\\
&=\frac{e^{x}-e^{-1}}{e^{x}+e^{-1}} \cdot \frac{e^{x}}{e^{x}}\\
&=\frac{e^{2 x}-1}{e^{2 x}+1}\\
t\left(e^{2 x}+1\right)&=e^{2 x}-1\\
(x-1) e^{2 x}+(x+1)&=0\\
e^{2 x}&= \frac{(t+1)}{(1-t)}\\
2x&=\ln \frac{(t+1)}{(1-t)}\\
x&= \frac{1}{2} \ln \frac{(t+1)}{(1-t)}\\
\end{aligned}
Hence
$$\tanh ^{-1} t=\frac{1}{2} \ln \left(\frac{1+t}{1-t}\right) \text { for }|t|<1$$
