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(B) Given,$\frac{d}{d x}\left[a 	an ^{-1} x+b \log \left(\frac{x-1}{x+1}ight)ight]=\frac{1}{x^4-1}$. Integrating both sides with respect to $x$,w

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$-1$

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(B) Given,$\frac{d}{d x}\left[a 	an ^{-1} x+b \log \left(\frac{x-1}{x+1}ight)ight]=\frac{1}{x^4-1}$. Integrating both sides with respect to $x$,we get: $a 	an ^{-1} x+b \log \left(\frac{x-1}{x+1}ight) = \int \frac{1}{x^4-1} dx$. We know that $\frac{1}{x^4-1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} ight]$. Thus,$\int \frac{1}{x^4-1} dx = \frac{1}{2} \int \frac{1}{x^2-1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$. Using standard integrals $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} ight|$ and $\int \frac{1}{x^2+1} dx = 	an^{-1} x$: $\int \frac{1}{x^4-1} dx = \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{x-1}{x+1} ight| ight) - \frac{1}{2} 	an^{-1} x = \frac{1}{4} \log \left| \frac{x-1}{x+1} ight| - \frac{1}{2} 	an^{-1} x$. Comparing this with $a 	an^{-1} x + b \log \left( \frac{x-1}{x+1} ight)$,we get $a = -\frac{1}{2}$ and $b = \frac{1}{4}$. Therefore,$a - 2b = -\frac{1}{2} - 2 \left( \frac{1}{4} ight) = -\frac{1}{2} - \frac{1}{2} = -1$.

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

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$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$ $\Rightarrow a-2 b$ is equal to