If $$\int \frac{dx}{x^3(1 + x^6)^{2/3}} = xf(x)(1 + x^6)^{1/3} + C$$, where C is a constant of integration, then the function $$f(x)$$ is equal to:

We start from the information that the antiderivative

$$\int \dfrac{dx}{x^{3}(1+x^{6})^{\;2/3}} = x\,f(x)\,(1+x^{6})^{1/3}+C,$$

where $$C$$ is the constant of integration. Because the integral of a function is an antiderivative, the derivative of the right-hand side must give back the integrand. Hence we write

$$\dfrac{d}{dx}\Bigl[x\,f(x)\,(1+x^{6})^{1/3}\Bigr] \;=\;\dfrac{1}{x^{3}(1+x^{6})^{\;2/3}}.$$

Now we differentiate the expression $$x\,f(x)\,(1+x^{6})^{1/3}$$ using the product rule. First state the product rule:

For two differentiable functions $$u(x)$$ and $$v(x)$$ we have $$\dfrac{d}{dx}[u(x)\,v(x)] = u'(x)\,v(x) + u(x)\,v'(x).$$

Here we have a product of three factors, so we apply the rule successively. Let

$$u_1 = x, \qquad u_2 = f(x), \qquad u_3 = (1+x^{6})^{1/3}.$$

Writing the derivative step by step:

$$\dfrac{d}{dx}\!\bigl[x\,f(x)\,(1+x^{6})^{1/3}\bigr] = f(x)(1+x^{6})^{1/3} + x\,f'(x)\,(1+x^{6})^{1/3} + x\,f(x)\,\dfrac{d}{dx}\!\bigl[(1+x^{6})^{1/3}\bigr].$$

We still need the derivative of $$(1+x^{6})^{1/3}.$$ Using the chain rule, stated as

$$\dfrac{d}{dx}[g(h(x))] = g'(h(x))\,h'(x),$$

we set $$g(t)=t^{1/3}$$ and $$h(x)=1+x^{6}.$$ Then

$$\dfrac{d}{dx}(1+x^{6})^{1/3} = \dfrac{1}{3}(1+x^{6})^{-2/3}\,(6x^{5}) = 2x^{5}(1+x^{6})^{-2/3}.$$

Substituting this back, the third term becomes

$$x\,f(x)\,\bigl[2x^{5}(1+x^{6})^{-2/3}\bigr] = 2x^{6}f(x)(1+x^{6})^{-2/3}.$$

Thus the whole derivative is

$$f(x)(1+x^{6})^{1/3} + x\,f'(x)(1+x^{6})^{1/3} + 2x^{6}f(x)(1+x^{6})^{-2/3}.$$

The first two terms contain the factor $$(1+x^{6})^{1/3},$$ and it is useful to express that with a common power of $$(1+x^{6}).$$ Note that

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

Replacing accordingly, we get

$$\bigl[f(x)+x\,f'(x)\bigr](1+x^{6})^{-2/3}(1+x^{6}) + 2x^{6}f(x)(1+x^{6})^{-2/3}.$$

The factor $$(1+x^{6})^{-2/3}$$ is common, so we factor it out:

$$\Bigl[(1+x^{6})\bigl(f(x)+x\,f'(x)\bigr)+2x^{6}f(x)\Bigr]\,(1+x^{6})^{-2/3}.$$

This entire derivative must equal the integrand, which we also write with the same power of $$(1+x^{6}):$$

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

Because the factor $$(1+x^{6})^{-2/3}$$ appears on both sides, we cancel it and obtain the differential equation

$$(1+x^{6})\bigl(f(x)+x\,f'(x)\bigr)+2x^{6}f(x) = x^{-3}.$$

Expanding the left-hand side completely:

$$(1+x^{6})\bigl(f+x\,f'\bigr)+2x^{6}f = f + x\,f' + x^{6}f + x^{7}f' + 2x^{6}f.$$

Combine like terms:

$$f + 3x^{6}f + x\,f' + x^{7}f' = x^{-3}.$$

Group the terms containing $$f$$ and $$f':$$

$$\bigl(1+3x^{6}\bigr)f + \bigl(x+x^{7}\bigr)f' = x^{-3}.$$

To solve this first-order linear equation, it is convenient to look for a power function that fits the pattern suggested by the answer choices. Suppose

$$f(x)=A\,x^{-k},$$

where $$A$$ and $$k$$ are constants. Differentiate this trial form:

$$f'(x)=\dfrac{d}{dx}\bigl(Ax^{-k}\bigr)= -kA\,x^{-k-1}.$$

Substitute $$f$$ and $$f'$$ into the differential equation:

$$\bigl(1+3x^{6}\bigr)(A\,x^{-k}) + \bigl(x+x^{7}\bigr)\bigl(-kA\,x^{-k-1}\bigr)=x^{-3}.$$

Simplify each part. First term:

$$\bigl(1+3x^{6}\bigr)A\,x^{-k}=A\,x^{-k}+3A\,x^{6-k}.$$

Second term:

$$\bigl(x+x^{7}\bigr)\bigl(-kA\,x^{-k-1}\bigr)= -kA\bigl(x\,x^{-k-1}+x^{7}x^{-k-1}\bigr) = -kA\bigl(x^{-k}+x^{6-k}\bigr).$$

Add them:

$$A\,x^{-k}+3A\,x^{6-k}-kA\,x^{-k}-kA\,x^{6-k} = A(1-k)\,x^{-k}+A(3-k)\,x^{6-k}.$$

This sum must equal $$x^{-3}.$$ For the equality to hold for all $$x,$$ each power of $$x$$ on the left must match a corresponding power on the right, or its coefficient must be zero. The right-hand side has only the term $$x^{-3},$$ so we want

1. One of the exponents on the left to be $$-3;$$ 2. The other term to vanish by having zero coefficient.

Take $$k=3.$$ Then

• The exponent $$-k$$ becomes $$-3,$$ matching the desired power. • The exponent $$6-k$$ becomes $$+3,$$ which we do not want; therefore its coefficient must be zero, i.e. $$3-k=0.$$ With $$k=3$$ indeed $$3-k=0,$$ so the unwanted term disappears.

With $$k=3$$ the remaining coefficient is

$$A(1-k)=A(1-3)=-2A.$$

This must equal the coefficient on the right, namely $$1.$$ Thus

$$-2A=1 \quad\Longrightarrow\quad A=-\dfrac{1}{2}.$$

Therefore

$$f(x)=A\,x^{-k}= -\dfrac{1}{2}\,x^{-3}= -\dfrac{1}{2x^{3}}.$$

This matches Option B.

Hence, the correct answer is Option B.