In the expansion of log_e frac{1}{1 - x - x^2 | vedclass

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(B) Given expression: $f(x) = \log_e \left[ \frac{1}{1 - x - x^2 + x^3} \right]$Factorizing the denominator: $1 - x - x^2 + x^3 = (1 - x) - x^2(1 - x)

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$1$

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(B) Given expression: $f(x) = \log_e \left[ \frac{1}{1 - x - x^2 + x^3} \right]$Factorizing the denominator: $1 - x - x^2 + x^3 = (1 - x) - x^2(1 - x) = (1 - x)(1 - x^2) = (1 - x)(1 - x)(1 + x) = (1 - x)^2(1 + x)$So,$f(x) = \log_e \left[ (1 - x)^{-2}(1 + x)^{-1} \right]$Using logarithmic properties: $f(x) = -2 \log_e(1 - x) - \log_e(1 + x)$Using the series expansions $\log_e(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$ and $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$$f(x) = -2(-x - \frac{x^2}{2} - \dots) - (x - \frac{x^2}{2} + \dots)$$f(x) = (2x + x^2 + \dots) - (x - \frac{x^2}{2} + \dots)$$f(x) = (2 - 1)x + (1 + 0.5)x^2 + \dots = x + 1.5x^2 + \dots$The coefficient of $x$ is $1$.

In the expansion of $\log_e \frac{1}{1 - x - x^2 + x^3}$,the coefficient of $x$ is

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