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

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

& \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{aligned}$$ Now we can write equation $(i)$ as $I=\int{-\dfrac{1}{2}\dfrac{{{t}^{2}}dt}{{{\left( {{t}^{3}} \right)}^{\dfrac{2}{3}}}}}$ $$\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{aligned} & \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{aligned}$$ Comparing both sides, we get $f(x)=-\dfrac{1}{2{{x}^{3}}}$ **So, the correct answer is “Option D”.** **Note:** The student must notice about putting the value of ${{t}^{3}}$ that we will put $$\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.