Question

The integral $$ \int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)} \, dx $$ is equal to: 

• A. \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C \)
• B. \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/2} + C \)
• C. \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right] + C \)
• D. \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{3} + C \)

Answer 1

The Correct Option is A

The integral to be evaluated is:

\(\int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)} \, dx\)

To solve this integral, we first analyze the components of the numerator and the denominator.

Consider the term within the inverse tangent function:

1. \(\frac{x^3 + 1}{x^3} = 1 + \frac{1}{x^3}\)

This simplification aids in the integration process.

Next, we perform a substitution:

1. \(t = \tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)\)

The differential of \(t\) is calculated as:

1. \(dt = \frac{d}{dx}\left(\tan^{-1}\left(1 + \frac{1}{x^3}\right)\right) \, dx = \frac{-3}{x^4(1 + (1 + \frac{1}{x^3})^2)} \, dx\)

Further simplification is achieved by noting that:

1. \((x^8 - x^2) is\ contained\ in\ the\ derivati\)

This substitution strategy enables effective integration.

Upon applying the substitution, we obtain:

\(\int f(t) \, dt = \log_e(t) + C\)

Substitute back the original expression for \(t\):

\(= \log_e\left(\tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)\right) + C\)

By applying properties of exponents and inverse tangents to align terms, we get:

\(= \log_e\left[\left(\tan^{-1}\left( \frac{x^3 + 1}{x^3} \right)\right)^{1/3}\right] + C\)

The final solution is:

\(\log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C\)