Question

If \[\int{\dfrac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{2}{3}}}}}=xf(x){{\left( 1+{{x}^{^{6}}} \right)}^{^{\dfrac{1}{3}}}}+c\] 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: To solve this question we have to make some arrangement of terms and use substitution as-
\[\int{\dfrac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{2}{3}}}}=\int{\dfrac{dx}{{{x}^{3}}{{\left( {{x}^{6}} \right)}^{\dfrac{2}{3}}}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}}}}\]
 \[=\int{\dfrac{dx}{{{x}^{3}}.{{x}^{4}}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}}}=\int{\dfrac{dx}{{{x}^{7}}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}}}\]
Put \[1+\dfrac{1}{{{x}^{6}}}={{t}^{3}}\] .

Complete step by step answer:
Let us consider
  \[I=\int{\dfrac{dx}{{{x}^{3}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{2}{3}}}}}\]
Now, let us take ${{x}^{6}}$ common from ${{(1+{{x}^{6}})}^{\dfrac{2}{3}}}$ . Then we will have
 $I=\int{\dfrac{dx}{{{x}^{3}}.{{\left( {{x}^{6}} \right)}^{\dfrac{2}{3}}}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}}}$
 \[\Rightarrow I=\int{\dfrac{dx}{{{x}^{3}}.{{x}^{4}}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}}}\]
  $\Rightarrow I=\int{\dfrac{dx}{{{x}^{7}}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}}}..............(i)$
 Let \[1+\dfrac{1}{{{x}^{6}}}={{t}^{3}}\] .
Differentiating, we get
  \[\begin{align}
  & \text{ }0+\left( -\dfrac{6}{{{x}^{7}}} \right)dx=3{{t}^{2}}dt \\
 & \Rightarrow -\dfrac{6}{{{x}^{7}}}dx=3{{t}^{2}}dt \\
 & \Rightarrow \dfrac{dx}{{{x}^{7}}}=-\dfrac{3}{6}{{t}^{2}}dt \\
 & \Rightarrow \dfrac{dx}{{{x}^{7}}}=-\dfrac{1}{2}{{t}^{2}}dt \\
\end{align}\]
Now we can write equation $(i)$ as
   $I=\int{-\dfrac{1}{2}\dfrac{{{t}^{2}}dt}{{{\left( {{t}^{3}} \right)}^{\dfrac{2}{3}}}}}$
  \[\Rightarrow I=-\dfrac{1}{2}\int{\dfrac{{{t}^{2}}dt}{{{t}^{2}}}}\]
  $\Rightarrow I=-\dfrac{1}{2}\int{dt}$
We know that $\int{dz=z+c}$ , where $c$ is a constant of integration.
  $\therefore I=-\dfrac{1}{2}t+c$
As we have put ${{t}^{3}}=\left( 1+\dfrac{1}{{{x}^{6}}} \right)$ , therefore
 $I=-\dfrac{1}{2}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{1}{3}}}+c$
Now it is given that
 \[\begin{align}
  & \text{ }\int{\dfrac{dx}{{{x}^{3}}{{(1+{{x}^{6}})}^{\dfrac{2}{3}}}}=xf(x)}{{(1+{{x}^{6}})}^{\dfrac{1}{3}}}+c \\
 & \Rightarrow I=xf(x){{(1+{{x}^{6}})}^{\dfrac{1}{3}}}+c \\
 & \Rightarrow -\dfrac{1}{2}{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{1}{3}}}+c=xf(x){{(1+{{x}^{6}})}^{\dfrac{1}{3}}}+c \\
 & \Rightarrow -\dfrac{1}{2}\dfrac{{{\left( 1+{{x}^{6}} \right)}^{\dfrac{1}{3}}}}{{{({{x}^{6}})}^{\dfrac{1}{3}}}}+c=xf(x){{(1+{{x}^{6}})}^{\dfrac{1}{3}}}+c \\
 & \Rightarrow -\dfrac{1}{2}\dfrac{1}{{{x}^{2}}}{{\left( 1+{{x}^{6}} \right)}^{\dfrac{1}{3}}}+c=xf(x){{(1+{{x}^{6}})}^{\dfrac{1}{3}}}+c \\
\end{align}\]
Comparing both sides, we get
 $f(x)=-\dfrac{1}{2{{x}^{3}}}$

So, the correct answer is “Option D”.

Note: The student must notice about putting the value of ${{t}^{3}}$ that we will put \[\left( 1+\dfrac{1}{{{x}^{6}}} \right)={{t}^{3}}\] , not \[{{\left( 1+\dfrac{1}{{{x}^{6}}} \right)}^{\dfrac{2}{3}}}={{t}^{3}}\] . Then you might get wrong or confused in finding the solution. Also, you must remember the integration of elementary functions.