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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) $\dfrac{2}{5}{{\left( 1-\dfrac{1}{{{x}^{3}}} \right)}^{\dfrac{5}{4}}}+c$

Last updated date: 16th Jul 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.

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 $'dt'$