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: Understanding the Concept:
This integration involves a substitution technique. We notice that the derivative of \(\tan^{-1}(x + 1/x)\) is likely present in the rest of the integrand.
Step 2: Detailed Explanation:
Let \(f(x) = \tan^{-1}(x + 1/x)\).
Compute the derivative:
\[ \frac{df}{dx} = \frac{1}{1 + (x + 1/x)^2} \cdot \frac{d}{dx}(x + 1/x) \]
\[ \frac{df}{dx} = \frac{1}{1 + (x^2 + 2 + 1/x^2)} \cdot (1 - 1/x^2) \]
\[ \frac{df}{dx} = \frac{1}{x^2 + 3 + 1/x^2} \cdot \left(\frac{x^2 - 1}{x^2}\right) \]
Multiply numerator and denominator of the first term by \(x^2\):
\[ \frac{df}{dx} = \frac{x^2}{x^4 + 3x^2 + 1} \cdot \frac{x^2 - 1}{x^2} \]
\[ \frac{df}{dx} = \frac{x^2 - 1}{x^4 + 3x^2 + 1} \]

The original integral is:
\[ I = \int \frac{\frac{x^2 - 1}{x^4 + 3x^2 + 1}}{\tan^{-1}(x + 1/x)} dx \]
Let \(t = \tan^{-1}(x + 1/x)\). Then \(dt = \frac{x^2 - 1}{x^4 + 3x^2 + 1} dx\).
Substituting these into the integral:
\[ I = \int \frac{1}{t} dt = \ln |t| + C \]
Substitute \(t\) back:
\[ I = \ln |\tan^{-1}(x + 1/x)| + C \]
Step 3: Final Answer:
The result is \(\ln |\tan^{-1}(x + 1/x)| + C\).