Answer

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**Hint:**Here the problem is based on integration with an unknown constant and a function on x which could be easily found by appropriate simplification and introducing needed terms. For that we must know some basic integral and differential formulas which are:

$\dfrac{d}{{dx}}{x^n} = n.{x^{n - 1}}$ and

$\dfrac{d}{{dx}}{(f(x))^n} = n.{(f(x))^{n - 1}}.f'(x)$

**Complete step by step answer:**

Step 1: Given $\int {\dfrac{{dx}}{{{x^3}{{(1 + {x^6})}^{\dfrac{2}{3}}}}} = f(x){{(1 + {x^{ - 6}})}^{\dfrac{1}{3}}} + C} $ ………………….(1)

The term which looks almost similar on LHS and RHS of equation (1) is ${(1 + {x^6})^{\dfrac{2}{3}}}$ on LHS and ${(1 + {x^{ - 6}})^{\dfrac{1}{3}}}$ on RHS. So

Let us name $t = {(1 + {x^{ - 6}})^{\dfrac{1}{3}}}$ . Then if we take its derivative with respect to the variable x we get,

$\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{3}{(1 + {x^{ - 6}})^{\dfrac{1}{3} - 1}}.( - 6{x^{ - 6 - 1}})$

$ = \dfrac{1}{3}{(1 + {x^{ - 6}})^{ - \dfrac{2}{3}}}.( - 6{x^{ - 7}})$

$ = ( - 2){(1 + {x^{ - 6}})^{ - \dfrac{2}{3}}}{x^{ - 7}}$ ………………….(2)

Step 2: Rearranging equation (2) to make it similar to equation (1), we get

$\Rightarrow ( - \dfrac{1}{2})dt = \dfrac{{dx}}{{{{(1 + {x^{ - 6}})}^{\dfrac{2}{3}}}{x^7}}}$ …………….(3)

RHS of obtained form is similar to the denominator of LHS in equation (1) with a difference in the power of x.

Step 3: Thus let us see whether we can make it similar or not.

We have ${x^3}{(1 + {x^6})^{\dfrac{2}{3}}} = {x^3}{x^4}{(1 + {x^{ - 6}})^{\dfrac{2}{3}}}$

$ = {x^7}{(1 + {x^{ - 6}})^{\dfrac{2}{3}}}$

Thus equation 1 changes to,

$\Rightarrow \int {\dfrac{{dx}}{{{x^7}{{(1 + {x^{ - 6}})}^{\dfrac{2}{3}}}}} = f(x){{(1 + {x^{ - 6}})}^{\dfrac{1}{3}}} + C} $ ………………….(4)

Step 4: Substituting equation (3) in LHS of equation (4), we get

\[\Rightarrow \int {\dfrac{{dx}}{{{x^7}{{(1 + {x^{ - 6}})}^{\dfrac{2}{3}}}}} = \int { - \dfrac{1}{2}dt = - \dfrac{1}{2}t + C} } \] where $t = {(1 + {x^{ - 6}})^{\dfrac{1}{3}}}$

Thus comparing with Equation (1), we have $f(x) = - \dfrac{1}{2}$ which is option A.

**Therefore, option (A) is correct. $f(x) = - \dfrac{1}{2}$**

**Note:**

In such problem solving we use integration by substitution which is used only when the derivative of RHS is present in the integrand part on LHS. In this method we substitute the RHS term under consideration as a single variable and take its derivative and substitute it in the LHS and solve. Similarly there are various other integration methods like Integration by parts which is given by the formula,

$\int {f(x).g(x)dx = f(x).\int {g(x) - \int {(f'(x)\int {g(x)dx)} } } } $ where f(x) and g(x) are two single variable functions.

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