Question:medium
The integral \(\int_{0}^{1} \cot^{-1}(1 + x + x^2) dx\) is equal to:
The integral \(\int_{0}^{1} \cot^{-1}(1 + x + x^2) dx\) is equal to:
Updated On: Apr 10, 2026
- \(2 \tan^{-1} 2 + \frac{1}{2} \log_e (\frac{5}{4}) + \frac{\pi}{2}\)
- \(2 \tan^{-1} 2 + \frac{1}{2} \log_e (\frac{5}{4}) - \frac{\pi}{2}\)
- \(2 \tan^{-1} 2 - \frac{1}{2} \log_e (\frac{5}{4}) + \frac{\pi}{2}\)
- \(2 \tan^{-1} 2 - \frac{1}{2} \log_e (\frac{5}{4}) - \frac{\pi}{2}\)
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The Correct Option is D
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