A) \[\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}+4}{2x} \right)+c\]
B) \[\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{2\sqrt{2x}} \right)+c\]
C) \[\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}+4}{2\sqrt{2x}} \right)+c\]
D) \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{2x} \right)+c\]
Correct Answer: B
Solution :
\[\int{\frac{{{x}^{2}}+4}{{{x}^{4}}+16}dx}\] \[=\int{\frac{1+\frac{4}{{{x}^{2}}}}{{{x}^{2}}+\frac{16}{{{x}^{2}}}}dx=\int{\frac{d\left( x-\frac{4}{x} \right)}{{{\left( x-\frac{4}{x} \right)}^{2}}+8}}}\] \[=\int{\frac{dt}{{{t}^{2}}+{{(2\sqrt{2})}^{2}}}}\], where\[t=x-\frac{4}{x}\] \[=\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{2\sqrt{2}x} \right)+c\]
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