A) \[{{\log }_{e}}({{x}^{2}}+1)\]
B) \[x{{\tan }^{-1}}x\]
C) \[\frac{{{\log }_{e}}({{x}^{2}}+1)}{2}\]
D) \[\frac{1}{2}x{{\tan }^{-1}}x\]
Correct Answer: C
Solution :
\[f(x)=\frac{x}{1+{{x}^{2}}}\], \[\therefore \,\,\,I=\int_{{}}^{{}}{f(x)}=\int_{{}}^{{}}{\frac{x}{1+{{x}^{2}}}\,dx}\] Put \[1+{{x}^{2}}=t\Rightarrow 2x\,dx=dt\Rightarrow x\,dx=dt/2\] \[\therefore \,\,\,I=\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{t}=\frac{1}{2}\log t+c}\]; \[I=\frac{1}{2}\log (1+{{x}^{2}})+c\].
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