A) \[{{\tan }^{-1}}(2{{x}^{2}}-1)+c\]
B) \[{{\tan }^{-1}}\frac{{{x}^{2}}+1}{x}+c\]
C) \[{{\sin }^{-1}}\left( x-\frac{1}{x} \right)+c\]
D) \[{{\tan }^{-1}}{{x}^{2}}+c\]
E) \[{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{x} \right)+c\]
Correct Answer: E
Solution :
Let \[I=\int{\frac{{{x}^{2}}+1}{{{x}^{4}}-{{x}^{2}}+1}}dx\] \[=\int{\frac{1+\frac{1}{{{x}^{2}}}}{{{x}^{2}}+\frac{1}{{{x}^{2}}}-1}}dx\] \[=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+1}}dx\] Let \[x-\frac{1}{x}=t\]and \[\left( 1+\frac{1}{{{x}^{2}}} \right)dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{{{t}^{2}}+1}}={{\tan }^{-1}}t+c\] \[={{\tan }^{-1}}\left( x-\frac{1}{x} \right)+c\] \[={{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{x} \right)+c\]
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