MHT CET · Maths · Definite Integration
The value of \(\int_0^1 \tan ^{-1}\left(1-x+x^2\right) \mathrm{d} x\) is
- A \(\frac{\pi}{2}-\log 2\)
- B \(\frac{\pi}{2}+\log 2\)
- C \(\log 2\)
- D 0
Answer & Solution
Correct Answer
(C) \(\log 2\)
Step-by-step Solution
Detailed explanation
\(I = \int_0^1 \tan ^{-1}\left(1-x+x^2\right) \mathrm{d} x\) Using \(\tan^{-1}(y) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{y}\right)\): \(I = \int_0^1 \left(\frac{\pi}{2} - \tan^{-1}\left(\frac{1}{1-x+x^2}\right)\right) \mathrm{d} x\)
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