A) 1
B) -1
C) 0
D) 2
Correct Answer: B
Solution :
Given, \[\frac{d}{dx}\left[ a{{\tan }^{-1}}x+b\log \left( \frac{x-1}{x+1} \right) \right]=\frac{1}{{{x}^{4}}-1}\] On integrating both sides, we get \[a{{\tan }^{-1}}x+b\log \left( \frac{x-1}{x+1} \right)\] \[=\frac{1}{2}\int{\left[ \frac{1}{{{x}^{2}}-1}-\frac{1}{{{x}^{2}}+1} \right]}\,dx\] \[\Rightarrow \] \[a{{\tan }^{-1}}x+b\log \left( \frac{x-1}{x+1} \right)\] \[=\frac{1}{4}\log \left( \frac{x-1}{x+1} \right)-\frac{1}{2}{{\tan }^{-1}}x\] \[\Rightarrow \] \[a=-\frac{1}{2},\] \[b=\frac{1}{4}\] \[\therefore \] \[a-2b=-\frac{1}{2}-2\left( \frac{1}{4} \right)=-1\]
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