Question:
$\displaystyle \int_{0}^{1} \left[\tan^{-1}\left(\frac{1}{1+x+x^2+x^3}\right) + \tan^{-1}(1+x+x^2+x^3)\right] dx$ is equal to:
$\displaystyle \int_{0}^{1} \left[\tan^{-1}\left(\frac{1}{1+x+x^2+x^3}\right) + \tan^{-1}(1+x+x^2+x^3)\right] dx$ is equal to:
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Look for inverse trig pair identities to simplify integrals instantly.
Updated On: Apr 24, 2026
- $\frac{\pi}{4}$
- $2\pi$
- $\frac{\pi}{2}$
- $1$
- $4\pi$
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The Correct Option is C
Solution and Explanation
Concept:
• Identity: \[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}, \quad x>0 \]
Step 1: Apply identity
\[ \tan^{-1}\left(\frac{1}{f(x)}\right) + \tan^{-1}(f(x)) = \frac{\pi}{2} \]
Step 2: Simplify integrand
\[ = \frac{\pi}{2} \]
Step 3: Integrate
\[ \int_{0}^{1} \frac{\pi}{2} dx = \frac{\pi}{2}(1-0) \] \[ = \frac{\pi}{2} \] Final Conclusion:
Option (C)
• Identity: \[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}, \quad x>0 \]
Step 1: Apply identity
\[ \tan^{-1}\left(\frac{1}{f(x)}\right) + \tan^{-1}(f(x)) = \frac{\pi}{2} \]
Step 2: Simplify integrand
\[ = \frac{\pi}{2} \]
Step 3: Integrate
\[ \int_{0}^{1} \frac{\pi}{2} dx = \frac{\pi}{2}(1-0) \] \[ = \frac{\pi}{2} \] Final Conclusion:
Option (C)
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