(C) We are given the equation $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$ for $|x| < 1$ ...$(i)$We know the standard logarithmic definition

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(C) We are given the equation $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$ for $|x| We know the standard logarithmic definition of the inverse hyperbolic tangent function is $\tanh ^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii)Comparing equation $(i)$ and equation (ii),we can equate the coefficients of the logarithmic term.Therefore,$a = \frac{1}{2}$.

Class 11 Mathematics Exponential and Logarithmic Series Logarithmic series

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If $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$,$|x| < 1$,then $a$ is equal to

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A
$1$
B
$2$
C
$\frac{1}{2}$
D
$\frac{1}{4}$

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Exponential and Logarithmic Series

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$\frac{m - n}{m + n} + \frac{1}{3}\left( \frac{m - n}{m + n} \right)^3 + \frac{1}{5}\left( \frac{m - n}{m + n} \right)^5 + \dots \infty = $

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