A) \[-\frac{{{({{x}^{4}}+1)}^{1/4}}}{x}+c\]
B) \[\frac{{{({{x}^{4}}+1)}^{1/4}}}{x}+c\]
C) zero
D) none of the above
Correct Answer: A
Solution :
Let \[I=\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}{{({{x}^{4}}+1)}^{3/4}}}}\] \[=\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}.{{x}^{3}}{{\left( 1+\frac{1}{{{x}^{4}}} \right)}^{3/4}}}}\] Put \[1+{{x}^{-4}}=t\] \[\Rightarrow \] \[\frac{-4}{{{x}^{5}}}dx=dt\] \[\therefore \] \[I=-\frac{1}{4}\int_{{}}^{{}}{\frac{dt}{{{t}^{3/4}}}}\] \[=-\frac{1}{4}\int_{{}}^{{}}{{{t}^{-3/4}}dt=-\frac{1}{4}\frac{{{t}^{1/4}}}{1/4}}+c\] \[=-{{\left( 1+\frac{1}{{{x}^{4}}} \right)}^{1/4}}+c\] \[=\frac{-{{({{x}^{4}}+1)}^{1/4}}}{x}+c\]
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