Question

If we are given an integral expression as $\int{\dfrac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{2}{3}}}}}=xf\left( x \right){{\left( 1+{{x}^{6}} \right)}^{\dfrac{1}{3}}}+C$ where, $C$ is a constant of integration, then the function $f\left( x \right)$ is equal to \[\]
A.$\dfrac{-1}{6{{x}^{3}}}$\[\]
B.$\dfrac{3}{{{x}^{2}}}$\[\]
C.$\dfrac{-1}{2{{x}^{2}}}$\[\]
D.$\dfrac{-1}{2{{x}^{3}}}$\[\]

Answer 1

Hint: We proceed from the left hand side and take ${{x}^{6}}$common from the bracket. We substitute ${{t}^{3}}=\dfrac{1}{{{x}^{6}}}+1$ and find $dx$ in terms of $t,dt$ also substitute in the integrand. We integrate with respect to $t$ and put back $t$ in terms of $x$. We multiply and divide $x$ and then compare the resultant expression with expression at the right hand side to get $f\left( x \right)$.\[\]

Complete step-by-step solution
We are given in the question an equation whose left hand side has an integral and the right hand side has functional equation as
\[\int{\dfrac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{2}{3}}}}}=xf\left( x \right){{\left( 1+{{x}^{6}} \right)}^{\dfrac{1}{3}}}+C\]
If we shall integrate the left hand side and try to express the result similar to the expression at the right hand side we may get the required function$f\left( x \right)$. We proceed from left hand side,
\[\Rightarrow \int{\dfrac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{2}{3}}}}}\]
We take ${{x}^{6}}$ common from the bracket in the denominator to get,
\[\begin{align}
  & \Rightarrow \int{\dfrac{dx}{{{x}^{3}}{{\left( {{x}^{6}}\left( \dfrac{1}{{{x}^{6}}}+1 \right) \right)}^{\dfrac{2}{3}}}}} \\
 & \Rightarrow \int{\dfrac{dx}{{{x}^{3}}{{\left( {{x}^{6}} \right)}^{\dfrac{2}{3}}}{{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{2}{3}}}}} \\
\end{align}\]
We use the exponential identity of power raised to another power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}},a\ne 0$ for $a=x,m=6,n=\dfrac{2}{3}$ in the above step to have
\[\begin{align}
  & \Rightarrow \int{\dfrac{dx}{{{x}^{3}}{{x}^{6\times \dfrac{2}{3}}}{{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{2}{3}}}}} \\
 & \Rightarrow \int{\dfrac{dx}{{{x}^{3}}{{x}^{4}}{{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{2}{3}}}}} \\
\end{align}\]
We use the exponential identity of power product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ for $a=x,m=3,n=4$ in the above step to have,
 \[\begin{align}
  & \Rightarrow \int{\dfrac{dx}{{{x}^{3+4}}{{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{2}{3}}}}} \\
 & \Rightarrow \int{\dfrac{dx}{{{x}^{7}}{{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{2}{3}}}}}.........\left( 1 \right) \\
\end{align}\]
We can substitute ${{t}^{3}}=\dfrac{1}{{{x}^{6}}}+1$ in the above step since integration is independent of change in variable and for that we also need $dx$ and rest of the expression in $x$ in terms of $t$. We differentiate ${{t}^{3}}=\dfrac{1}{{{x}^{6}}}+1$ both side with respect to $x$ to have,
\[\begin{align}
  & \dfrac{d}{dx}{{t}^{3}}=\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{6}}}+1 \right)=\dfrac{d}{dx}\left( {{x}^{-6}}+1 \right) \\
 & \Rightarrow 3{{t}^{2}}\dfrac{dt}{dx}=-6{{x}^{-7}}+0 \\
 & \Rightarrow dx=\dfrac{3{{t}^{2}}}{-6{{x}^{-7}}} dt \\
\end{align}\]

We put ${{t}^{3}}=\dfrac{1}{{{x}^{6}}}+1$ and $dx=\dfrac{3{{t}^{2}}}{-6{{x}^{-7}}}$ in (1) to have
\[\begin{align}
  & \Rightarrow \int{\dfrac{\dfrac{3{{t}^{2}}}{-6{{x}^{-7}}}dt}{{{x}^{7}}{{\left( {{t}^{3}} \right)}^{\dfrac{2}{3}}}}} \\
 & \Rightarrow \int{\dfrac{3{{t}^{2}}dt}{-6{{x}^{-7}}\cdot {{x}^{7}}{{\left( {{t}^{3}} \right)}^{\dfrac{2}{3}}}}} \\
\end{align}\]
We use the exponential identity of power product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ for $a=x,m=-7,n=7$ and exponential identity of power raised to another power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}},a\ne 0$ for $a=t,m=3,n=\dfrac{2}{3}$ in the above step to have
\[\begin{align}
  & \Rightarrow \int{\dfrac{{{t}^{2}}dt}{-2{{x}^{-7+7}}{{t}^{3\times \dfrac{2}{3}}}}} \\
 & \Rightarrow \dfrac{-1}{2}\int{\dfrac{{{t}^{2}}dt}{{{x}^{0}}{{t}^{2}}}} \\
 & \Rightarrow \dfrac{-1}{2}\int{dt} \\
 & \Rightarrow \dfrac{-1}{2}t+C \\
\end{align}\]
We take cube root of both side of the equation ${{t}^{3}}=\dfrac{1}{{{x}^{6}}}+1$ and find $t={{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{1}{3}}}$. We put $t$ in the above step to have,
\[\begin{align}
  & \Rightarrow \dfrac{-1}{2}{{\left( \dfrac{1}{{{x}^{6}}}+1 \right)}^{\dfrac{1}{3}}}+C \\
 & \Rightarrow \dfrac{-1}{2}{{\left( \dfrac{{{x}^{6}}+1}{{{x}^{6}}} \right)}^{\dfrac{1}{3}}}+C \\
 & \Rightarrow \dfrac{-1}{2}\dfrac{{{\left( {{x}^{6}}+1 \right)}^{\dfrac{1}{3}}}}{{{\left( {{x}^{6}} \right)}^{\dfrac{1}{3}}}}+C \\
\end{align}\]
We use exponential identity of power raised to another power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}},a\ne 0$ for $a=x,m=6,n=\dfrac{1}{3}$ in the above step to have
\[\Rightarrow \dfrac{-1}{2}\dfrac{{{\left( {{x}^{6}}+1 \right)}^{\dfrac{1}{3}}}}{{{x}^{2}}}+C\]
Let us multiply and divide $x$ with the expression to make it comparable with the expression at right hand side $xf\left( x \right){{\left( 1+{{x}^{6}} \right)}^{\dfrac{1}{3}}}$. We have,
\[\begin{align}
  & \Rightarrow \dfrac{-1}{2}x\dfrac{{{\left( {{x}^{6}}+1 \right)}^{\dfrac{1}{3}}}}{{{x}^{2}}\cdot x}+C \\
 & \Rightarrow \dfrac{-1}{2{{x}^{3}}}x{{\left( {{x}^{6}}+1 \right)}^{\dfrac{1}{3}}}+C \\
 & \Rightarrow x\left( \dfrac{-1}{2{{x}^{3}}} \right){{\left( {{x}^{6}}+1 \right)}^{\dfrac{1}{3}}}+C=xf\left( x \right){{\left( 1+{{x}^{6}} \right)}^{\dfrac{1}{3}}} \\
\end{align}\]
We compare both sides and have the required function as $f\left( x \right)=\dfrac{-1}{2{{x}^{3}}}$. So the correct option is D.

Note: We note that the function is not defined for $x=0$ and the question assumes that. The method of integration we used here is called t-substitution or integration by substitution. We have also used the formula standard integral $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1},n\ne -1$ and the standard differentiation formula $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ frequently in this problem.