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Simplify the following integral:
\[\int{\dfrac{{{\left( {{x}^{4}}-x \right)}^{\dfrac{1}{4}}}}{{{x}^{5}}}dx}\]
(a) \[\dfrac{2}{5}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{-5}{4}}}+c\]
(b) \[\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{-5}{4}}}+c\]
(c) \[\dfrac{4}{15}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c\]
(d) \[\dfrac{2}{5}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c\]

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Last updated date: 29th Feb 2024
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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 \['dt'\]