If y = 2x^2 - 1,then left[ frac{1}{y} + frac{ | vedclass

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(B) We know the logarithmic series expansion: $\frac{1}{2} \ln \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} + \dots$ for $|z| <

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$\frac{1}{2} \left[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots \right]$

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(B) We know the logarithmic series expansion: $\frac{1}{2} \ln \left( \frac{1+z}{1-z} \right) = z + \frac{z^3}{3} + \frac{z^5}{5} + \dots$ for $|z| Let $z = \frac{1}{y}$. Then the given expression is $\frac{1}{2} \ln \left( \frac{1 + 1/y}{1 - 1/y} \right) = \frac{1}{2} \ln \left( \frac{y+1}{y-1} \right)$.Substitute $y = 2x^2 - 1$:$\frac{1}{2} \ln \left( \frac{2x^2 - 1 + 1}{2x^2 - 1 - 1} \right) = \frac{1}{2} \ln \left( \frac{2x^2}{2x^2 - 2} \right) = \frac{1}{2} \ln \left( \frac{x^2}{x^2 - 1} \right) = -\frac{1}{2} \ln \left( \frac{x^2 - 1}{x^2} \right) = -\frac{1}{2} \ln \left( 1 - \frac{1}{x^2} \right)$.Using the expansion $\ln(1-u) = -(u + \frac{u^2}{2} + \frac{u^3}{3} + \dots)$ where $u = \frac{1}{x^2}$:$-\frac{1}{2} [ -(\frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots) ] = \frac{1}{2} [ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \dots ]$.

If $y = 2x^2 - 1$,then $\left[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5} + \dots \right]$ is equal to

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