A) \[2[x{{\tan }^{-1}}x-\log (1+{{x}^{2}})]+C\]
B) \[2[x{{\tan }^{-1}}x+\log (1+{{x}^{2}})]+C\]
C) \[2x{{\tan }^{-1}}x+\log (1+{{x}^{2}})+C\]
D) \[2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C\]
Correct Answer: D
Solution :
Given, when \[x>0\] \[=\int{{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)dx}\] \[\left[ \begin{align} & \because \,\,x>0 \\ & \therefore \,\,2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \\ \end{align} \right]\] \[=\int{2{{\tan }^{-1}}\,x\,\,dx}\] \[=2\int{\underset{II}{\mathop{1}}\,.\,\underset{I}{\mathop{ta{{n}^{-1}}}}\,\,x\,\,dx}\] \[=2\left\{ x.{{\tan }^{-1}}x-\int{\frac{1}{1+{{x}^{2}}}.x\,dx} \right\}\] \[=2x{{\tan }^{-1}}x-\int{\frac{2x}{1+{{x}^{2}}}dx}\] \[=2x\,{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C\]
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