Q73 / 120 · VITEEE 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
- A1
- Correct answerB$-1$
- C0
- D2
Solution
$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^{4}-1}$
On integrating both sides, we get
$$
\begin{array}{l}
a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) \\
\quad=\frac{1}{2} \int\left[\frac{1}{x^{2}-1}-\frac{1}{x^{2}+1}\right] d x \\
\Rightarrow a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) \\
\quad=\frac{1}{4} \log \left(\frac{x-1}{x+1}\right)-\frac{1}{2} \tan ^{-1} x \\
\Rightarrow a=-\frac{1}{2}, b=\frac{1}{4} \\
\therefore a-2 b=-\frac{1}{2}-2\left(\frac{1}{4}\right)=-1
\end{array}
$$
Hence, the correct answer is option: 'B'.