A) \[\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{2x\sqrt{2}} \right)+C\]
B) \[\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{2\sqrt{2}} \right)+C\]
C) \[\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{x\sqrt{2}} \right)+C\]
D) None of these
Correct Answer: A
Solution :
Let\[I=\int{\frac{{{x}^{2}}+4}{{{x}^{4}}+16}dx}\] \[=\int{\frac{1+\frac{4}{{{x}^{2}}}}{{{x}^{2}}+\frac{16}{{{x}^{2}}}}dx}\] \[=\int{\frac{1+\frac{4}{{{x}^{2}}}}{{{\left( x-\frac{4}{x} \right)}^{2}}+8}dx}\] Putting \[x-\frac{4}{x}=t,\] So that \[\left( 1+\frac{4}{{{x}^{2}}} \right)dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{{{t}^{2}}+{{(2\sqrt{2})}^{2}}}}\] \[=\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{t}{2\sqrt{2}} \right)+C\] \[\Rightarrow \] \[I=\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{x-\frac{4}{x}}{2\sqrt{2}} \right)+C\] \[=\frac{1}{2\sqrt{2}}{{\tan }^{-1}}\left( \frac{{{x}^{2}}-4}{2x\sqrt{2}} \right)+C\]
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