A) \[x{{\tan }^{-1}}x+\frac{1}{2}\log (1+{{x}^{2}})\]
B) \[x{{\tan }^{-1}}x-\frac{1}{2}\log (1+{{x}^{2}})\]
C) \[(x-1){{\tan }^{-1}}x\]
D) \[x{{\tan }^{-1}}x-\log (1+{{x}^{2}})\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{{{\tan }^{-1}}x\,dx}=x{{\tan }^{-1}}x-\int_{{}}^{{}}{\frac{x}{1+{{x}^{2}}}\,dx+c}\] \[=x{{\tan }^{-1}}x-\frac{1}{2}\log (1+{{x}^{2}})+c.\] Note : Students should remember this question as a formula.
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