A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B) \[-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{1}{1+{{x}^{2}}}\]
D) \[-\frac{1}{1+{{x}^{2}}}\]
Correct Answer: C
Solution :
Let \[I=\frac{d}{dx}\left\{ \begin{align} & {{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right) \\ & -{{\tan }^{-1}}\,\left( \frac{4x-4{{x}^{3}}}{1-6{{x}^{2}}+{{x}^{4}}} \right) \\ \end{align} \right\}\] Put \[x=\tan \theta \] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}x\] \[\therefore \] \[I=\frac{d}{dx}\{{{\tan }^{-1}}(\tan 2\theta )+{{\tan }^{-1}}(\tan 3\theta )\] \[-{{\tan }^{-1}}(\tan \,4\theta )\}\] \[=\frac{d}{dx}(2\theta +3\theta -4\theta )\] \[=\frac{d}{dx}(\theta )\] \[=\frac{d}{dx}({{\tan }^{-1}}x)\] \[=\frac{1}{1+{{x}^{2}}}\]
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