Question:
\( \frac{d}{dx} \left[ a \tan^{-1} x + b \log\left(\frac{x-1}{x+1}\right) \right] = \frac{1}{x^{4} - 1} \implies a - 2b \) is equal to
\( \frac{d}{dx} \left[ a \tan^{-1} x + b \log\left(\frac{x-1}{x+1}\right) \right] = \frac{1}{x^{4} - 1} \implies a - 2b \) is equal to
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Partial fraction decomposition: $\frac{1}{A \cdot B} = \frac{1}{B-A} (\frac{1}{A} - \frac{1}{B})$.
Updated On: Apr 10, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Integrate RHS
Integrate $\frac{1}{x^4 - 1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]$. $\int \frac{1}{x^4 - 1} dx = \frac{1}{2} \int \frac{1}{x^2-1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$. $= \frac{1}{2} \cdot \frac{1}{2} \log\left(\frac{x-1}{x+1}\right) - \frac{1}{2} \tan^{-1} x = \frac{1}{4} \log\left(\frac{x-1}{x+1}\right) - \frac{1}{2} \tan^{-1} x$.
Step 2: Match Coefficients
Comparing with $a \tan^{-1} x + b \log\left(\frac{x-1}{x+1}\right)$: $a = -1/2$ and $b = 1/4$.
Step 3: Calculate Result
$a - 2b = -1/2 - 2(1/4) = -1/2 - 1/2 = -1$.
Final Answer: (b)
Integrate $\frac{1}{x^4 - 1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]$. $\int \frac{1}{x^4 - 1} dx = \frac{1}{2} \int \frac{1}{x^2-1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$. $= \frac{1}{2} \cdot \frac{1}{2} \log\left(\frac{x-1}{x+1}\right) - \frac{1}{2} \tan^{-1} x = \frac{1}{4} \log\left(\frac{x-1}{x+1}\right) - \frac{1}{2} \tan^{-1} x$.
Step 2: Match Coefficients
Comparing with $a \tan^{-1} x + b \log\left(\frac{x-1}{x+1}\right)$: $a = -1/2$ and $b = 1/4$.
Step 3: Calculate Result
$a - 2b = -1/2 - 2(1/4) = -1/2 - 1/2 = -1$.
Final Answer: (b)
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