Key Concepts and Formulas

• Indefinite Integration: The process of finding the antiderivative of a function.
• Substitution Method: A technique used to simplify integrals by substituting a part of the integrand with a new variable.
• Power Rule for Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$ and C is the constant of integration.

Step-by-Step Solution

Step 1: Rewrite the integral to prepare for substitution. We are given the integral: $I = \int \frac{dx}{x^3(1 + x^6)^{2/3}}$ We want to manipulate the integrand to make a suitable substitution. We factor out $x^6$ from the term inside the parenthesis: $I = \int \frac{dx}{x^3(x^6(1/x^6 + 1))^{2/3}} = \int \frac{dx}{x^3(x^4(1/x^6 + 1)^{2/3})} = \int \frac{dx}{x^7(1/x^6 + 1)^{2/3}}$

Step 2: Perform the substitution. Let $t = \frac{1}{x^6} + 1$. Then, differentiating with respect to $x$, we get: $\frac{dt}{dx} = -\frac{6}{x^7}$ $dt = -\frac{6}{x^7} dx$ $\frac{dx}{x^7} = -\frac{1}{6} dt$

Step 3: Substitute into the integral and evaluate. Substituting $t$ and $dt$ into the integral, we have: $I = \int \frac{dx}{x^7(1/x^6 + 1)^{2/3}} = \int \frac{-1/6 dt}{t^{2/3}} = -\frac{1}{6} \int t^{-2/3} dt$ Now, using the power rule for integration: $I = -\frac{1}{6} \frac{t^{1/3}}{1/3} + C = -\frac{1}{6} \cdot 3 \cdot t^{1/3} + C = -\frac{1}{2} t^{1/3} + C$

Step 4: Substitute back for $t$. Replacing $t$ with $\frac{1}{x^6} + 1$, we get: $I = -\frac{1}{2} \left(\frac{1}{x^6} + 1\right)^{1/3} + C = -\frac{1}{2} \left(\frac{1 + x^6}{x^6}\right)^{1/3} + C = -\frac{1}{2} \frac{(1 + x^6)^{1/3}}{x^2} + C$

Step 5: Rewrite the result in the desired form. We want to express the integral in the form $xf(x)(1 + x^6)^{1/3} + C$. So, we multiply and divide by $x$: $I = -\frac{1}{2} \frac{(1 + x^6)^{1/3}}{x^2} \cdot \frac{x}{x} + C = x \left(-\frac{1}{2x^3}\right) (1 + x^6)^{1/3} + C$ Thus, we can identify $f(x) = -\frac{1}{2x^3}$.

Step 6: Check the given answer. The problem states that the correct answer is $f(x) = \frac{3}{x^2}$. This contradicts our calculation. Let's differentiate $xf(x)(1+x^6)^{1/3}$ using the product rule: $\frac{d}{dx} \left[ x \left( \frac{3}{x^2} \right) (1+x^6)^{1/3} \right] = \frac{d}{dx} \left[ \frac{3}{x} (1+x^6)^{1/3} \right] = 3 \frac{d}{dx} \left[ x^{-1} (1+x^6)^{1/3} \right]$ $= 3 \left[ -x^{-2} (1+x^6)^{1/3} + x^{-1} \frac{1}{3} (1+x^6)^{-2/3} (6x^5) \right] = 3 \left[ -\frac{(1+x^6)^{1/3}}{x^2} + \frac{2x^4}{(1+x^6)^{2/3}} \right]$ $= 3 \left[ \frac{-(1+x^6) + 2x^6}{x^2 (1+x^6)^{2/3}} \right] = 3 \left[ \frac{-1 - x^6 + 2x^6}{x^2 (1+x^6)^{2/3}} \right] = 3 \left[ \frac{x^6 - 1}{x^2 (1+x^6)^{2/3}} \right]$ This does not seem to simplify to $\frac{1}{x^3(1+x^6)^{2/3}}$.

Let's consider the correct answer $f(x) = \frac{3}{x^2}$. Then, we are given $\int \frac{dx}{x^3 (1+x^6)^{2/3}} = x \frac{3}{x^2} (1+x^6)^{1/3} + C = \frac{3}{x} (1+x^6)^{1/3} + C$ Differentiating both sides with respect to $x$, we have $\frac{1}{x^3 (1+x^6)^{2/3}} = \frac{d}{dx} \left[ \frac{3}{x} (1+x^6)^{1/3} + C \right] = 3 \frac{d}{dx} \left[ x^{-1} (1+x^6)^{1/3} \right]$ $= 3 \left[ -x^{-2} (1+x^6)^{1/3} + x^{-1} \frac{1}{3} (1+x^6)^{-2/3} (6x^5) \right] = 3 \left[ -\frac{(1+x^6)^{1/3}}{x^2} + \frac{2x^4}{(1+x^6)^{2/3}} \right]$ $= 3 \left[ \frac{-(1+x^6) + 2x^6}{x^2 (1+x^6)^{2/3}} \right] = 3 \left[ \frac{x^6 - 1}{x^2 (1+x^6)^{2/3}} \right]$ This is clearly not equal. There must be an error in the given solution. I calculated $f(x) = -\frac{1}{2x^3}$.

Common Mistakes & Tips

• Carefully track negative signs during substitution.
• Remember to substitute back to the original variable after integration.
• Always double-check your work by differentiating the final result.

Summary

We are given the integral $\int \frac{dx}{x^3(1 + x^6)^{2/3}}$ and the form of the solution $xf(x)(1 + x^6)^{1/3} + C$. By using substitution and the power rule for integration, we find that $f(x) = -\frac{1}{2x^3}$. The original answer provided in the question is incorrect.

Final Answer

The final answer is \boxed{-\frac{1}{2x^3}}, which corresponds to option (C).