\[\int_{{}}^{{}}{\frac{{{({{x}^{4}}-x)}^{1/4}}}{{{x}^{5}}}\ dx}\] is equal to

\[\frac{4}{5}{{\left( 1-\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]

\[\frac{4}{15}{{\left( 1-\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]

\[\frac{4}{15}{{\left( 1+\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]

\[\int_{{}}^{{}}{\frac{{{({{x}^{4}}-x)}^{1/4}}}{{{x}^{5}}}\ dx}\] is e

1.	\[\int_{{}}^{{}}{\frac{{{({{x}^{4}}-x)}^{1/4}}}{{{x}^{5}}}\ dx}\] is equal to
A.	\[\frac{4}{15}{{\left( 1-\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]
B.	\[\frac{4}{5}{{\left( 1-\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]
C.	\[\frac{4}{15}{{\left( 1+\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]
D.	None of these
Answer» B. \[\frac{4}{5}{{\left( 1-\frac{1}{{{x}^{3}}} \right)}^{5/4}}+c\]

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\[\int_{{}}^{{}}{\frac{{{({{x}^{4}}-x)}^{1/4}}}{{{x}^{5}}}\ dx}\] is equal to

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