Evaluate: log_e sqrt{frac{1+x}{1-x}} | vedclass

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Question

(A) We know that $\log_e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} \right)$.Using the logarithmic expansion $\log_e \left( \f

Vedclass · Accepted Answer

$x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$

Vedclass · Answer

(A) We know that $\log_e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} ight)$.Using the logarithmic expansion $\log_e \left( \frac{1+x}{1-x} ight) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots ight)$ for $|x| Substituting this into the expression:$\frac{1}{2} 	imes 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots ight) = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$Thus,the correct option is $A$.

Evaluate: $\log_e \sqrt{\frac{1+x}{1-x}}$

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