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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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Verified Answer
The correct answer is:
$\frac{1}{2}$
$\tanh ^{-1} x=a \log \left(\frac{1+x}{1-x}\right),|x| < 1$ ...(i)
But $\tanh ^{-1} x=\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii)
From Eqs. (i) and (ii), we get
$a=\frac{1}{2}$
But $\tanh ^{-1} x=\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii)
From Eqs. (i) and (ii), we get
$a=\frac{1}{2}$
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