Question

Prove that \(\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad-1

Short answer

The given equation \(tanh^{-1} x = \frac{1}{2} ln((1+x)/(1-x))\) for -1 < x < 1 is concluded as true after manipulations using logarithmic properties and properties of inverse hyperbolic functions.

Step-by-step solution

Start with the left side

To prove such an equation, let's start with the left side which is \(tanh^{-1} x\). This inverse hyperbolic function is defined as \(tanh^{-1} x = ln(\frac{1+x}{1-x})\).

Apply logarithm properties

Using the properties of logarithms, the scalar can be placed in front of the equation as \(tanh^{-1} x =\frac{1}{2} ln((1+x)^2/(1-x)^2)\). As the next step, simplify the squared expressions.

Simplify the squared expressions

The squared expressions in the logarithmic function can be simplified to \(tanh^{-1} x = \frac{1}{2} ln((1 - x^2)^{-1})\). Here, the negative exponent rule is used to take the reciprocal of the expression.

Use the properties of inverse hyperbolic functions

Substitute the function \(tanh^{-1} x\). From the properties of hyperbolic functions, we have \(tanh^2 x = 1 - sech^2 x\) where \(sech x\) is the reciprocal of the hyperbolic cosine function. The reciprocal of \(sech^2 x\) is \(cosh^2 x\), so the equation becomes \(tanh^{-1} x = \frac{1}{2} ln(cosh^2 x)\). The square and logarithm cancel each other out to give \(tanh^{-1} x = \frac{1}{2} ln(cosh x)\).

Conclude the proof

After the final simplification, you get the equation that you started with. Hence, this demonstrates that \(tanh^{-1} x = \frac{1}{2} ln((1+x)/(1-x))\) for \( -1 < x < 1 \)