A) \[f(x)={{x}^{2}}+1\]
B) \[f(x)=\frac{{{x}^{2}}+1}{2x}\]
C) \[f(x)=\frac{{{x}^{2}}+1}{x}\]
D) \[f(x)=\frac{1}{2}({{x}^{2}}+1)\]
Correct Answer: C
Solution :
\[I=\int_{{}}^{{}}{\frac{({{x}^{2}}-1)dx}{({{x}^{4}}+3{{x}^{2}}+1){{\tan }^{-1}}\left( \frac{{{x}^{2}}+1}{x} \right)}}\] (Dividing Num. and Den. by x2) \[I=\int_{{}}^{{}}{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)dx}{\left( {{x}^{2}}+3+\frac{1}{{{x}^{2}}} \right){{\tan }^{-1}}\left( x+\frac{1}{x} \right)}}=\int_{{}}^{{}}{\frac{dt}{\left( {{t}^{2}}+1 \right){{\tan }^{-1}}t}}\]\[\left( \text{where}\,\text{t}=x+\frac{1}{x}\Rightarrow dt=\left( 1-\frac{1}{{{x}^{2}}} \right)dx \right)\] \[=\log |{{\tan }^{-1}}t|+C=\log \left| {{\tan }^{-1}}\left( \frac{{{x}^{2}}+1}{x} \right) \right|+C\] \[\Rightarrow f(x)=\frac{{{x}^{2}}+1}{x}\]
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