Question

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

• A. $\ln|\tan^{-1}(x+1/x)| + C$
• B. $\tan^{-1}(x+1/x) + C$
• C. $\frac{1}{2} (\tan^{-1}(x+1/x))^2 + C$
• D. $\ln|x^4+3x^2+1| + C$

Hint:

When you see $x^2 \pm 1$ in the numerator and $x^4 + kx^2 + 1$ in the denominator, look for substitutions involving $(x \pm 1/x)$.

Answer 1

The Correct Option is A

Step 1: Concept
Use substitution method. Let $t = \tan^{-1}(x + \frac{1}{x})$.

Step 2: Meaning
Differentiate $t$ with respect to $x$:
$dt = \frac{1}{1 + (x + \frac{1}{x})^2} \cdot \frac{d}{dx}(x + \frac{1}{x}) dx$
$dt = \frac{1}{1 + x^2 + 2 + \frac{1}{x^2}} \cdot (1 - \frac{1}{x^2}) dx$

Step 3: Analysis
Simplify the derivative:
$dt = \frac{1}{\frac{x^4 + 3x^2 + 1}{x^2}} \cdot \frac{x^2 - 1}{x^2} dx = \frac{x^2 - 1}{x^4 + 3x^2 + 1} dx$
The entire numerator and the quadratic part of the denominator (except for $\tan^{-1}$) is exactly $dt$.

Step 4: Conclusion
The integral becomes $\int \frac{1}{t} dt = \ln|t| + C$.
Substituting back for $t$: $\ln|\tan^{-1}(x + \frac{1}{x})| + C$. Final Answer: (A)