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Inverse Hyperbolic Tangent — Definition, Formula & Examples

The inverse hyperbolic tangent, written artanh⁡(x)\operatorname{artanh}(x) or tanh⁡−1(x)\tanh^{-1}(x), is the function that reverses the hyperbolic tangent — it returns the value yy such that tanh⁡(y)=x\tanh(y) = x.

For x∈(−1,1)x \in (-1, 1), the inverse hyperbolic tangent is defined as artanh⁡(x)=12ln⁡ ⁣(1+x1−x)\operatorname{artanh}(x) = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right), where ln⁡\ln denotes the natural logarithm. It is the unique real number yy satisfying tanh⁡(y)=x\tanh(y) = x.

Key Formula

artanh⁡(x)=12ln⁡ ⁣(1+x1−x),∣x∣<1\operatorname{artanh}(x) = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right), \quad |x| < 1
Where:
  • xx = Input value, restricted to the open interval (-1, 1)
  • ln⁡\ln = Natural logarithm (base e)

How It Works

To evaluate artanh⁡(x)\operatorname{artanh}(x), substitute xx into the logarithmic formula 12ln⁡ ⁣(1+x1−x)\frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right). The domain is strictly between −1-1 and 11 because the fraction inside the logarithm must be positive. As x→±1x \to \pm 1, the output diverges to ±∞\pm\infty. In calculus, the derivative is ddxartanh⁡(x)=11−x2\frac{d}{dx}\operatorname{artanh}(x) = \frac{1}{1-x^2}, which connects this function to partial-fraction integration techniques.

Worked Example

Problem: Evaluate artanh⁡(0.5)\operatorname{artanh}(0.5) using the logarithmic formula.
Step 1: Substitute x=0.5x = 0.5 into the formula.
artanh⁡(0.5)=12ln⁡ ⁣(1+0.51−0.5)=12ln⁡ ⁣(1.50.5)\operatorname{artanh}(0.5) = \frac{1}{2}\ln\!\left(\frac{1 + 0.5}{1 - 0.5}\right) = \frac{1}{2}\ln\!\left(\frac{1.5}{0.5}\right)
Step 2: Simplify the fraction inside the logarithm.
=12ln⁡(3)= \frac{1}{2}\ln(3)
Step 3: Compute the numerical value using ln⁡(3)≈1.0986\ln(3) \approx 1.0986.
≈12(1.0986)=0.5493\approx \frac{1}{2}(1.0986) = 0.5493
Answer: artanh⁡(0.5)=12ln⁡3≈0.5493\operatorname{artanh}(0.5) = \frac{1}{2}\ln 3 \approx 0.5493

Why It Matters

The inverse hyperbolic tangent appears when integrating 11−x2\frac{1}{1-x^2} in calculus courses, providing a cleaner alternative to partial fractions. It also arises in physics (relativistic velocity addition) and statistics (the Fisher z-transformation used in correlation analysis).

Common Mistakes

Mistake: Plugging in x=1x = 1 or x=−1x = -1 and expecting a finite result.
Correction: The domain is the open interval (−1,1)(-1, 1). At the endpoints, 1+x1−x\frac{1+x}{1-x} becomes 00 or undefined, so artanh⁡(±1)\operatorname{artanh}(\pm 1) does not exist as a real number.