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**Hint:**We solve this problem by converting the integral part as power of some variable by assuming the variable as some function of \['x'\] that is

\[t=f\left( x \right)\]

Then we differentiate that assumption to get the value of \['dx'\] in terms of \['dt'\] so as to substitute in the given integral to get in the standard form that is \[\int{{{t}^{n}}dt}\]

We use the standard formula of integration that is

\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c\]

By using the above formula we get the required value of integral.

**Complete step-by-step answer:**

Let us assume that the given integral as

\[\Rightarrow I=\int{\dfrac{{{\left( {{x}^{4}}-x \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx}\]

Now by taking the common term out from the numerator and cancelling with denominator we get

\[\begin{align}

& \Rightarrow I=\int{\dfrac{{{\left( {{x}^{4}} \right)}^{\dfrac{1}{4}}}{{\left( 1-\dfrac{x}{{{x}^{4}}} \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx} \\

& \Rightarrow I=\int{\dfrac{{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{1}{4}}}}{{{x}^{4}}}dx}..........equation(i) \\

\end{align}\]

Now, let us assume that the value inside the bracket as some other variable that is

\[\Rightarrow t=1-\dfrac{1}{{{x}^{3}}}\]

Now, by differentiating with respect to \['x'\] on both sides we get

\[\Rightarrow \dfrac{dt}{dx}=\dfrac{d}{dx}\left( 1 \right)-\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{3}}} \right)\]

We know that the standard formulas of differentiation as

\[\begin{align}

& \dfrac{d}{dx}\left( \text{constant} \right)=0 \\

& \dfrac{d}{dx}\left( {{x}^{n}} \right)=n.{{x}^{n-1}} \\

\end{align}\]

By using these formulas to above equation we get

\[\begin{align}

& \Rightarrow \dfrac{dt}{dx}=0-\left( -3.{{x}^{-3-1}} \right) \\

& \Rightarrow dt=\dfrac{3}{{{x}^{4}}}dx \\

& \Rightarrow \dfrac{1}{{{x}^{4}}}dx=\dfrac{dt}{3} \\

\end{align}\]

Now, by substituting the required values in equation (i) we get

\[\begin{align}

& \Rightarrow I=\int{{{\left( t \right)}^{\dfrac{1}{4}}}\left( \dfrac{dt}{3} \right)} \\

& \Rightarrow I=\dfrac{1}{3}\int{{{t}^{\dfrac{1}{4}}}dt} \\

\end{align}\]

We know that the standard formula of integration that is

\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c\]

By using this formula in above equation we get

\[\begin{align}

& \Rightarrow I=\dfrac{1}{3}\left( \dfrac{{{t}^{\dfrac{1}{4}+1}}}{\dfrac{1}{4}+1} \right)+c \\

& \Rightarrow I=\dfrac{1}{3}\left( \dfrac{4}{5}{{t}^{\dfrac{5}{4}}} \right) \\

& \Rightarrow I=\dfrac{4}{15}{{t}^{\dfrac{5}{4}}} \\

\end{align}\]

Now, by substituting the value of \['t'\] in terms of \['x'\] in above equation we get

\[\Rightarrow I=\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c\]

Therefore the value of given integral is

\[\therefore \int{\dfrac{{{\left( {{x}^{4}}-x \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx}=\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c\]

**So, the correct answer is “Option C”.**

**Note:**Students may make mistakes in converting the integral of \['x'\] to integral of some other variable.

We have the value of integration as

\[\Rightarrow I=\int{\dfrac{{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{1}{4}}}}{{{x}^{4}}}dx}\]

Then we assume that

\[\Rightarrow t=1-\dfrac{1}{{{x}^{3}}}\]

After this assumption we need to differentiate the above equation to get the value as

\[\Rightarrow \dfrac{1}{{{x}^{4}}}dx=\dfrac{dt}{3}\]

But, some students miss this differentiation and take the value as\[dx=dt\] which will be wrong. We need to differentiate the above equation to get \['dx'\] in terms of \['dt'\]

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