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 given integral 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 the integral, observe the structure of both the numerator and denominator:

Notice the expression inside the inverse tangent function:

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

This transformation will help simplify the integration process.

Next, let's substitute:

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

The derivative of \(t\) is:

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

We can further simplify using:

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

This equips us to use the substitution effectively to find the integral.

Upon applying it, we can see:

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

Re-substitute the value of \(t\):

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

Matching it with the power to clear denominators and inverse tangent properties gives:

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

Therefore, the correct answer is:

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

Answer 2

Given:
\(I = \int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \tan^{-1}\left(x^3 + \frac{1}{x^3}\right)} dx.\)

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

Then:  
\(dt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \left(3x^2 - \frac{3}{x^4}\right) dx.\)

Simplifying:
\(dt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \frac{3x^6 - 3}{x^4} dx.\)

\(dt = \frac{x^6 - 1}{x^{12} + 3x^6 + 1} dx.\)

Rewriting the integral:
\(I = \frac{1}{3} \int \frac{dt}{t} = \frac{1}{3} \ln|t| + C.\)

Substituting back:
\(I = \frac{1}{3} \ln\left|\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right| + C.\)

Simplifying further:
\(I = \ln\left(\tan^{-1}\left(x^3 + \frac{1}{x^3}\right)\right)^{1/3} + C.\)

The correct option is (A) : \( \log_e\left[\tan^{-1}\left(\frac{x^3 + 1}{x^3}\right)\right]^{1/3} + C \)