Key Concepts and Formulas

• Substitution Method for Integration: $\int f(g(x))g'(x) \, dx = \int f(u) \, du$, where $u = g(x)$ and $du = g'(x) \, dx$.
• Derivative of Inverse Tangent: $\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \frac{du}{dx}$.
• Integral of 1/x: $\int \frac{1}{x} \, dx = \ln|x| + C$.

Step-by-Step Solution

Step 1: Define the integral and prepare for substitution.

• We are given the integral: $I = \int \frac{(x^8 - x^2)}{(x^{12} + 3x^6 + 1) \tan^{-1}(x^3 + \frac{1}{x^3})} \, dx$
• We recognize that the derivative of $x^3 + \frac{1}{x^3}$ might be related to the numerator. This motivates a substitution involving $\tan^{-1}(x^3 + \frac{1}{x^3})$.

Step 2: Perform the substitution.

• Let $t = \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)$.
• Then, we need to find $dt$. Using the chain rule and the derivative of the inverse tangent function: $\frac{dt}{dx} = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \frac{d}{dx} \left(x^3 + \frac{1}{x^3}\right)$ $\frac{dt}{dx} = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \left(3x^2 - \frac{3}{x^4}\right)$ $\frac{dt}{dx} = \frac{1}{1 + x^6 + 2 + \frac{1}{x^6}} \cdot \left(3x^2 - \frac{3}{x^4}\right)$ $\frac{dt}{dx} = \frac{1}{x^6 + 3 + \frac{1}{x^6}} \cdot \left(3x^2 - \frac{3}{x^4}\right)$ $\frac{dt}{dx} = \frac{1}{\frac{x^{12} + 3x^6 + 1}{x^6}} \cdot \frac{3x^6 - 3}{x^4}$ $\frac{dt}{dx} = \frac{x^6}{x^{12} + 3x^6 + 1} \cdot \frac{3(x^6 - 1)}{x^4}$ $\frac{dt}{dx} = \frac{3x^2(x^6 - 1)}{x^{12} + 3x^6 + 1}$
• Therefore, $dt = \frac{3x^2(x^6 - 1)}{x^{12} + 3x^6 + 1} \, dx = \frac{3(x^8 - x^2)}{x^{12} + 3x^6 + 1} \, dx$.

Step 3: Rewrite the integral in terms of t.

• From the previous step, we have $\frac{dt}{3} = \frac{(x^8 - x^2)}{x^{12} + 3x^6 + 1} \, dx$.
• Substituting this into the original integral, we get: $I = \int \frac{1}{\tan^{-1}(x^3 + \frac{1}{x^3})} \cdot \frac{(x^8 - x^2)}{x^{12} + 3x^6 + 1} \, dx = \int \frac{1}{t} \cdot \frac{dt}{3}$ $I = \frac{1}{3} \int \frac{1}{t} \, dt$

Step 4: Evaluate the simplified integral.

• The integral of $\frac{1}{t}$ is $\ln|t|$.
• Thus, $I = \frac{1}{3} \ln|t| + C$.

Step 5: Substitute back for t.

• Recall that $t = \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)$.
• Substituting this back into the expression for I, we get: $I = \frac{1}{3} \ln\left|\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right| + C$

Step 6: Rewrite the answer using logarithm properties.

• Using the property $a \ln(x) = \ln(x^a)$, we can rewrite the integral as: $I = \ln\left|\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right|^{\frac{1}{3}} + C$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when differentiating $x^3 + \frac{1}{x^3}$. Remember that $\frac{d}{dx}(x^{-3}) = -3x^{-4}$.
• Algebraic Simplification: The algebraic manipulation to arrive at the expression for dt can be tricky. Double-check each step.
• Absolute Value: Remember to include the absolute value inside the logarithm, as the argument of the logarithm must be positive. The range of $\tan^{-1}(x^3 + 1/x^3)$ does not guarantee a positive value.

Summary

We solved the integral using the substitution method. By recognizing the relationship between the derivative of $x^3 + \frac{1}{x^3}$ and the numerator of the integrand, we were able to simplify the integral to a basic logarithmic form. After evaluating the integral and substituting back, we arrived at the final answer.

The final answer is $\boxed{\ln\left|\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right|^{\frac{1}{3}} + C}$, which corresponds to option (A).