Step 1: Recall the definition of $\tanh^{-1} x$.
$\tanh x = \frac{e^x - e^{-x}}{e^x+e^{-x}}$. To find the inverse, let $y = \tanh x$, so $\frac{e^x-e^{-x}}{e^x+e^{-x}} = y$.
Step 2: Solve for $x$ in terms of $y$.
$e^x-e^{-x} = y(e^x+e^{-x}) \Rightarrow e^x(1-y) = e^{-x}(1+y) \Rightarrow e^{2x} = \frac{1+y}{1-y}$ (valid when $|y|<1$).
Step 3: Take logarithm.
$2x = \log\frac{1+y}{1-y} \Rightarrow x = \frac{1}{2}\log\frac{1+y}{1-y}$.
Step 4: Write the inverse function.
Replacing $y$ with $x$: \[ \tanh^{-1} x = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right), \quad |x|<1 \]
Step 5: Match with the options.
Option 1 is $\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$. This matches.
Step 6: State the answer.
\[ \boxed{\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)} \]