A) \[\frac{1}{2}({{x}^{2}}+1){{\tan }^{-1}}x-\frac{1}{2}x+c\]
B) \[\frac{1}{2}({{x}^{2}}-1){{\tan }^{-1}}x-\frac{1}{2}x+c\]
C) \[\frac{1}{2}({{x}^{2}}+1){{\tan }^{-1}}x+\frac{1}{2}x+c\]
D) \[\frac{1}{2}({{x}^{2}}+1){{\tan }^{-1}}x-x+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{x\,.\,{{\tan }^{-1}}x\,dx=\frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}\int_{{}}^{{}}{\frac{{{x}^{2}}+1-1}{1+{{x}^{2}}}\,dx}}\] \[=\frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}x+\frac{1}{2}{{\tan }^{-1}}x+c\] \[=\frac{1}{2}{{\tan }^{-1}}x\,.\,({{x}^{2}}+1)-\frac{1}{2}x+c\].
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