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

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