A) \[\frac{1}{\sqrt{{{x}^{2}}-\frac{1}{{{x}^{2}}}}}+C\]
B) \[\frac{1}{\sqrt{\frac{1}{{{x}^{2}}}-{{x}^{2}}}}+C\]
C) \[\frac{1}{\sqrt{\frac{1}{{{x}^{2}}}+{{x}^{2}}}}+C\]
D) None of these
Correct Answer: D
Solution :
\[I=\int{\frac{1+{{x}^{4}}}{{{(1-{{x}^{4}})}^{3/2}}}dx=\int{\frac{{{x}^{3}}\left( x+\frac{1}{{{x}^{3}}} \right)dx}{{{(1-{{x}^{4}})}^{3/2}}}}}\] \[=\frac{\left( x+\frac{1}{{{x}^{3}}} \right)dx}{{{\left( \frac{1}{{{x}^{2}}}-{{x}^{2}} \right)}^{3/2}}}\] Let, \[\frac{1}{{{x}^{2}}}-{{x}^{2}}=t\] \[\Rightarrow \] \[\left( \frac{-2}{{{x}^{3}}}-2x \right)dx=dt\] \[\Rightarrow \] \[x+\frac{1}{{{x}^{3}}}dx=\frac{-1}{2}dt\] \[\Rightarrow \] \[t=-\frac{1}{2}\int{\frac{dt}{{{t}^{3/2}}}}=\frac{1}{\sqrt{t}}+C=\frac{1}{\sqrt{\frac{1}{{{x}^{2}}}-{{x}^{2}}}}+C\]
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