Question:
\( \int \frac{x^2 - 1}{(x^4 + 3x^2 + 1)\tan^{-1}\left(x + \frac{1}{x}\right)} \, dx \) is equal to
\( \int \frac{x^2 - 1}{(x^4 + 3x^2 + 1)\tan^{-1}\left(x + \frac{1}{x}\right)} \, dx \) is equal to
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If derivative of inner function appears in numerator → try substitution.
Updated On: Apr 23, 2026
- $\tan^{-1}\left(x + \frac{1}{x}\right) + C$
- $\cot^{-1}\left(x + \frac{1}{x}\right) + C$
- $\log\left(x + \frac{1}{x}\right) + C$
- $\log\left[\tan^{-1}\left(x + \frac{1}{x}\right)\right] + C$
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The Correct Option is D
Solution and Explanation
Concept:
Use substitution method with derivative matching.
Step 1: Let $t = \tan^{-1}\left(x + \frac{1}{x}\right)$.
Step 2: Differentiate $t$.
\[ \frac{dt}{dx} = \frac{1}{1+\left(x+\frac{1}{x}\right)^2} \cdot \left(1 - \frac{1}{x^2}\right) \] \[ = \frac{x^2 - 1}{x^2(1 + x^2 + \frac{1}{x^2})} \] \[ = \frac{x^2 - 1}{x^4 + 3x^2 + 1} \]
Step 3: Substitute in integral.
\[ \int \frac{dt}{t} \]
Step 4: Integrate.
\[ = \log |t| + C \]
Step 5: Substitute back.
\[ = \log\left[\tan^{-1}\left(x + \frac{1}{x}\right)\right] + C \] Conclusion:
Answer = Option (D)
Step 1: Let $t = \tan^{-1}\left(x + \frac{1}{x}\right)$.
Step 2: Differentiate $t$.
\[ \frac{dt}{dx} = \frac{1}{1+\left(x+\frac{1}{x}\right)^2} \cdot \left(1 - \frac{1}{x^2}\right) \] \[ = \frac{x^2 - 1}{x^2(1 + x^2 + \frac{1}{x^2})} \] \[ = \frac{x^2 - 1}{x^4 + 3x^2 + 1} \]
Step 3: Substitute in integral.
\[ \int \frac{dt}{t} \]
Step 4: Integrate.
\[ = \log |t| + C \]
Step 5: Substitute back.
\[ = \log\left[\tan^{-1}\left(x + \frac{1}{x}\right)\right] + C \] Conclusion:
Answer = Option (D)
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