A) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{2x} \right)+c\]
B) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{\sqrt{2x}} \right)+c\]
C) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{2\sqrt{x}} \right)+c\]
D) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{\sqrt{2}x} \right)+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{{{x}^{2}}+1}{{{x}^{4}}+1}}\,dx=\int_{{}}^{{}}{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}}\,dx=\int_{{}}^{{}}{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)\,dx}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}}\] Put \[x-\frac{1}{x}=t\Rightarrow \left( 1+\frac{1}{{{x}^{2}}} \right)\,dx=dt,\] then the required integral is \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{\sqrt{2}x} \right)+c\].
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