Question

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

Answer 1

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

& \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{aligned}$$ Now, let us assume that the value inside the bracket as some other variable that is $$\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{aligned} & \dfrac{d}{dx}\left( \text{constant} \right)=0 \\\ & \dfrac{d}{dx}\left( {{x}^{n}} \right)=n.{{x}^{n-1}} \\\ \end{aligned}$$ By using these formulas to above equation we get $$\begin{aligned} & \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{aligned}$$ Now, by substituting the required values in equation (i) we get $$\begin{aligned} & \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{aligned}$$ We know that the standard formula of integration that is $$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$$ By using this formula in above equation we get $$\begin{aligned} & \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{aligned}$$ Now, by substituting the value of $$'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'$$