Q107 / 160 · AP EAMCET 2009 · Mathematics

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

  • A
    1
  • Correct answer
    B
    -1
  • C
    0
  • D
    2

Solution

Given, $\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$ On integrating both sides, we get $\begin{aligned} & a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) \\ & =\frac{1}{2} \int\left[\frac{1}{x^2-1}-\frac{1}{x^2+1}\right] d x \\ \Rightarrow \quad & a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) \\ & =\frac{1}{4} \log \left(\frac{x-1}{x+1}\right)-\frac{1}{2} \tan ^{-1} x \\ \Rightarrow \quad & a=-\frac{1}{2}, \quad b=\frac{1}{4} \\ \therefore \quad & a-2 b=-\frac{1}{2}-2\left(\frac{1}{4}\right)=-1 \end{aligned}$

Hence, the correct answer is option: 'B'.

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