Question

The value of $\int \frac{x^2 - 1}{(x^4 + 3x^2 + 1) \tan^{-1}(x + \frac{1}{x})} dx$ is:

• A. $\log | \tan^{-1}(x + 1/x) | + C$
• B. $\tan^{-1}(x + 1/x) + C$
• C. $\log(x + 1/x) + C$
• D. None of these

Hint:

For integrals with $x^4$ and $x^2$ terms, dividing by $x^2$ often reveals a $x \pm 1/x$ substitution.

Answer 1

The Correct Option is A

Step 1: Concept
Divide both the numerator and the denominator by $x^2$ to transform the integrand into a form suitable for substitution.

Step 2: Meaning
The integral becomes $\int \frac{1 - 1/x^2}{(x^2 + 3 + 1/x^2) \tan^{-1}(x + 1/x)} dx$. We can rewrite the quadratic term as $x^2 + 1/x^2 + 3 = (x + 1/x)^2 + 1$.

Step 3: Analysis
Let $t = \tan^{-1}(x + 1/x)$. Differentiating gives $dt = \frac{1}{1 + (x + 1/x)^2} \cdot (1 - 1/x^2) dx$. The numerator and the quadratic part of the denominator exactly match $dt$.

Step 4: Conclusion
The integral reduces to $\int \frac{1}{t} dt = \log |t| + C$. Substituting $t$ back gives $\log | \tan^{-1}(x + 1/x) | + C$. Final Answer: (A)