Question:
Let \(x\in \mathbb{R}\) and \(|x|\lt 1\). Then
\[
\tanh^{-1}x=
\]
Let \(x\in \mathbb{R}\) and \(|x|\lt 1\). Then
\[
\tanh^{-1}x=
\]
Show Hint
Remember the inverse hyperbolic tangent formula:
\[
\tanh^{-1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right),\quad |x|\lt 1.
\]
Updated On: Jun 24, 2026
- \(\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)\)
- \(\frac{1}{2}\log\left(\frac{1-x}{1+x}\right)\)
- \(\frac{1}{2}\log\left(x+\sqrt{1-x^2}\right)\)
- \(\frac{1}{2}\log\left(x-\sqrt{1-x^2}\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the standard formula.
For \(|x|\lt 1\), \[ \tanh^{-1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right) \]
Step 2: Match with the options.
The expression \[ \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) \] matches option \((1)\).
Step 3: Final conclusion.
Therefore, \[ \boxed{\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)} \]
For \(|x|\lt 1\), \[ \tanh^{-1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right) \]
Step 2: Match with the options.
The expression \[ \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) \] matches option \((1)\).
Step 3: Final conclusion.
Therefore, \[ \boxed{\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)} \]
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